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MANY  ADDITIONAL    EXAMPLES,   ILLUSTRATING 
THE    ALGEBRAIC    ANALYSIS. 


By  CHARLES   DAVIES,  LL.D.. 

AUTHOR  OF  A  FULL  COURSE  OF  MATHEMATICS, 


Vfr^,. 


-  I_ 


A.   S.    BARNES    &    COMPANY, 

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Copyright,  185'',  by  Chas.  Davies,  LL.D. 
Copyright,  1879,  by  A.  S.  Barnes  &  Co. 


PREFACE. 


A  wiDS  difference  of  opinion  is  known  to  exist  among  teachers 
in  regard  to  the  value  of  a  Key  to  any  mathematical  work,  and 
it  is  perhaps  yet  undecided  whether  a  Key  is  a  help  or  a 
hindrance. 

If  a  Key  is  designed  to  supersede  the  necessity  of  investigfw 
tion  and  labor  on  the  part  of  the  teacher;  to  present  to  his 
mind  every  combination  of  thought  which  ought  to  be  suggested 
by  a  problem,  and  to  permit  him  to  float  sluggishly  along  the 
current  of  ideas  developed  by  the  author,  it  would  certainly  do 
great  harm,  and  should  be  excluded  from  every  school. 

If,  on  the  contrary,  a  Key  is  so  constructed  as  to  suggest 
ideas,  both  in  regard  to  particular  questions  and  general  science, 
which  the  Text-book  might  not  impart;  if  it  develops  methods 
of  solution  too  particular  or  to-,  elaborate  to  find  a  place  in  the 
test ;  if  it  is  mainly  designed  to  lessen  the  mechanical  labor  of 
teaching,  rather  than  the  labor  of  study  and  investigation ;  i*. 
may,  in  the  hands  of  a  good  teacher,  prove  a  valuable  auxiliary. 

The    Key    to  Bourdon    is  intended    to  answer,    precisely,    thia 


-I  Q  O  /?  O  o 


IT  PREFACE. 

end.  The  principles  developed  in  the  text  are  explained  and 
Ulustrated  by  means  of  numerous  examples,  and  these  are  all 
wrought  in  the  Key  by  methods  which  accord  with  and  make 
evident  the  principles  themselves.  The  Key,  therefore,  not  only 
explains  the  various  questions,  but  is  a  commentary  on  the  text 
itself. 

Nothing  is  more  gratifying  to  an  ambitious  teacher  than  to 
push  forward  the  investigations  of  his  pupils  beyond  the  limits 
of  the  text  book.  To  aid  him  hi  an  undertaking  so  useful  to 
himself  and  to  them,  an  Appendix  has  been  added,  containing  a 
copious  collection  of  Practical  Examples.  Mauy  of  the  solutions 
are  quite  curious  and  in.-tructive ;  and  taken  in  connection  with 
those  embraced  in  the  Text,  form  a  full  and  complete  system  of 
Algebraic  Analysis. 

The  many  letters  which  I  have  received  from  Teachers  and 
Pupils,  in  regard  to  the  best  solutions  of  new  questions,  have 
suggested  the  desirableness  of  furnishing,  in  the  present  work, 
those  which  have  been  most  approved.  They  are  a  collection 
of  problems  that  have  special  values,  and  their  solutions  may  be 
studied  with  great  profit  by  every  one  seeking  mathematical 
knowledge. 

FisHKiLL  Landing,  ) 
July,  1S73.        \ 


INTRODUCTION. 


ALGEBRA. 


1.    On   an   analysis  of  the  subject  of  Algebra,  we      Aig«b«. 
think  it  will  appear  that  the  subject  itself  presents  no     DifficuUi«. 
serious  difficulties,  and  that  most  of  the  embarrassment 
which  is  experienced  by  the  pupil  in  gaining  a  knowl-      "°^J7" 
edge  of  its  principles,  as  well  as  in  their  applications, 
arises  from   not  attending    sufficiently   to   the   lanyuage      Langua«<^ 
or   signs  of  the   thoughts  which    are   combined   in   the 
reasonings.     At  the  hazard,  therefore,  of  being  a  little 
diffuse,    I   shall   begin   with    the   very  elements  of  the 
algebraic   language,   and   explain,    with    much    minute- 
ness, the  exact  signification  of  tlie  characters  that  stand 


Characters 
which  repre- 

for  the  quantities  which  are  the  subjects  of  the  analy-  sent  quantity 


sis  ;  and  also  of  those  signs  which  indicate  the  several 
operations  to  be  performed  on  the  quantities. 


Signs. 


2.  The  quantities  which  are  the  subjects  of  the  Huantitie* 
algebraic  analysis  may  be  divided  into  two  classes :  ^°^  divid«a 
those  which  are  known  or  given,  and  those  which  are 
unknown  or  sought.  The  known  are  uniformily  repre- 
sented by  the  first  letters  of  the  alphabet,  «,  b,  c,  rf, 
&c. ;  and  the  unknown  by  the  final  letters,  r,  y,  2, 
t',   w.   &c. 


How  repre- 
sented. 


6 


mTEODUCTION. 


May  be  in- 
creased or 
diminisheJ. 

Five  opera- 
tions 


Quantity  is  susceptible  of  "being  increased,  di- 
minished, and  measured  ;  and  there  are  six  operations 
which  can  be  performed  upon  a  quantity  that  will 
give  results  differing  from  the   quantity  itself,  viz.: 

1st.   To  add  it  to  itself  or  to  some  other  quantity; 

2d.    To  subtract  some  other  quantity  from  it; 

3d.    To  multiply  it  by  a  number; 

4th.  To  divide  it; 

5th.  To  raise  it  to  any  power; 

6th.  To  extract  a  root  of  it. 

The  cases  in  which  the  multiplier  or  divisor  is  1, 
are  of  course  excepted;  as  also  the  case  where  a 
root  is  to  be  extracted  of  1. 

4.  The  six  signs  which   denote   these   operations 
Elements      ^^.^  ^^^  ^,gjj  j^jiq^^jj  {q  ^g  repeated  here.     These,  with 

of  the  ^  " 

Algebr^iic       the  signs  of  equality  and  inequality,  tlie  letters  of  the 

language. 

alphabet  and  the  figures  which  are  employed,  make  up 
ts  words  and   ji,g    elements  of   the  algebraic    Janfruace.     The    words 

phrases  : 

and    phrases    of    the    algebraic,    like    those    of    every 
How  inter-     other    lanfjuaofe,  are    to    be    taken    in    connection   with 

preted. 

each  other,  and  are  not    to  be    interpreted  as  separate 
and  isolated  symbols 


First 

Becand 
Third. 

Fourth 
Fifth 

Exception. 
Signs. 


Symbols  of 
quantity. 


5.  The  symbols  of  quantity  are  designed  to  repre- 
sent quantity  in  general,  whether  abstract  or  concrete, 
whether  known  or  unknown  ;  and  the  signs  which  in- 
dicate the  operations  to  be  performed  on  the  quanti- 
ties are  to  be  interpreted  in  a  sense  equally  general. 
When  the  sign  plus  is  written,  it  indicate*  that  the 
eigns  pins  and  quantity  before  which  it  is  placed  is  to  be  added  to 
some  other    quantity  :  and    the  sign  minus    implies  the 


General. 


Esample?. 


INTROPrCTION.  < 

existence  of  a  minuend,  from  which  the  subtrahend  is 

to  be  taken.     One  thing  should  be  observed  in  regard  Signs  have  n* 

effect  on  tho 

to  the  signs  which   indicate  the  operations   that  are  to       nature  of 

t  7  I        1        I!       ^  quantity. 

be  performed  on  quantities,  viz. :  thev  do  not  at  all 
affict  or  change  the  nature  of  the  quantity  before  or 
lifter  'which  they  are  tvritten,  hnt  merely  indicate  u'hat 
is   to   be   done   ivith    the   quantity.     In  Algebra,  for  ex-     i^xamplet: 

In  Algebra. 

ample,  the  minus  sign  merely  indicates  that  the  quan- 
tity before  which  it  is  written  is  to  be  subtracted  from 
some  other  quantity;  and  in  Anal)tical  Geometry,  that    in  Analytical 

Geometry. 

the  line  before  which  it  falls  is  estimated  in  a  contrary 
direction  from  that  in  which  it  would  have  been  reck- 
oned, had  it  had  the  sign  plus  ;  but  in  neither  case  is 
the  nature  of  the  quantity  itself  different  from  what 
it   would   have   been   had  the   sign   been   plus. 

The    interpretation    of   the    language    of  Algebra    is   Tnterpretat.on 

of  the 

the  first  thing  to  which  the  attention  of  a  pupil  should      u.iguag* 

be  directed  ;  and  he  should  be  drilled  on  the  meaning 

and    import    of  the    symbols,    until    their    significations 

and   uses   are   as   familiar   as   the  sounds   and  combina-    '*'  "«<=««^"' 

tions   of    the   letters   of   the   alphabet. 

6     Beo-inning    with    the   elements   of  the    language,       lUt^menu 

ex;<lai.-iP.J 

let  any  number  or  quantity  be  designated  by  the  letter 

«,  and   let   it   be   required  to   add   this   letter  to   it.-elf 

and    find    the    result   or    sum.      The    additiDu    will    be 

expressed  by 

a  -\-  a  ^z  the  sum. 

But   how    is    the    sum    to   be    expressed?     By    simply    Significatie. 
reo'ardina  a  as  one  a,  or  la,  and   then  observing   that 
one  a  and  one  a,  make  tico  aV    or  2a :   hence. 


S  INTRODUCTION. 

a  -f-  a  =  2a;  . 

and  thus  we  place  a  figure  before  a  letter  to  indicate 
how  many  times   it  is  taken.     Such  figure  is  called  a 
Co-effieient      Co-effi,cienL 

Pwduct:  7.    The    product   of    several    numbers   is   indicated 

by  the  sign  of  multiplication,  or  by  simply  writing  the 
letters  which  represent  the  numbers  by  the  side  of 
each  other.      Thus, 


Iaw  indicated 


Factor. 


ay^bxcxdxf, 


or 


abcdf. 


indicates  the  continued  product  of  cr,  h,  c,  d,  and  _/, 
and  each  letter  is  called  a  factor  of  the  product : 
hence,  a  factor  of  a  product  is  one  of  the  multipliers 
which  produce   it.     Any  figure,  as  5,  written  before  a 

product,  as 

oabcdf, 


Co-efficient  of   is  the  co-eflScient  of  the  product,  and  shows  that  the 
•  P'   ^'^  •      product  is   taken   5   times. 

E;iiai  factors :         g^    jf    the    numbers   represented   by  a,  b,  c,  d,  and 
»hatthe        /,    were    equal    to    each    other,    they    would    each    be 
wmel       represented    by    a    single    letter    a,   and    the    product 
would    then    become 


How 
•xpr«utc 


axaxaxaxa  =  a~; 


that  is,  we  indicate  the  product  of  several  equal  fao 
tors  by  simply  writing  the  letter  once  and  placing  a 
figure  above  and  a  Tnle  at   the  right    )f  it,  to  indicate 


INTRODUCTION. 


9 


how  many  times  it  is  taken  as  a  factor.     The  figure     Exponent : 
60  written  is  called   an   exponent.     Hence,  on  exponent  where  writun. 
\g  a  simple  form  of    langituge    to  point  mtt  how  many 
equal  factors  are  employed. 


9.  The  division  of  one  quantity  by  another  is  indi- 
cated by  simply  writing  the  divisor  below  the  dividend, 
after  the  manner  of  a  fraction  ;  by  placing  it  on  the 
right  of  the  dividend  with  a  horizontal  line  and  two 
dots  between  them  ;  or  by  placing  it  on  the  right  with 
a  vertical  line  between  them :  thus  either  form  of 
expression : 


DiTiiioB : 

how 
expresEed 


a 


a, 


or 


Three  fonna 


mdicates  the  division  of  b  by  a. 

10.    The   extraction  of  a   root   is   indicated  by  the        Roots 

sign   -y/.     This  sign,  when  used  by  itself  indicates  the  how  indicate 
lowest   root,  viz.,  the  square   root.      If  any  other  root 
is   to  be  extracted,  as   the  third,  fourth,  fifth,  &c.,  the        index; 

figure  marking  the  degree  of  the  root  is  written  above  where  written 
and  at  the  left  of  the  sign  ;  as, 

^/    cube  root,  \/    ^fourth  root,  &c. 


The  figure  so  written,  is  called  the  Index  of  the  root. 
"We  have  thus  given  the  very  simple  and  general 
language  by  which  we  indicate  each  of  the  five 
operations  that  may  be  performed  on  an  algebraic 
quantity,  and  every  process  iit  Algebra  involves  one  or 
otlu'r  of  these  operations. 


Language  fsi 

the  five 
opera  tian* 


10  INTRODUCTION.       . 

MINUSSIGN. 

Algebraic  ^j     rpj^^   alsrc^braic    symbols    are    divided    into   two 

language  ■  '  ^  J 

classes    entirely    distinct    from     each     otlier — viz.,    the 

kow  diTiJe:     letters  that  are  used  to  designate  the   quantities  which 

are   the   subjects   of  the   science,  and    the   signs   v.hich 

are  employed  to  indicate  certain  operations   to  be  per- 

Algebraic       formed    on    those  quantities.      We    have    seen    that    all 

processes  :  ... 

the  algebraic  processes  are  comprised  under  addition, 
ihei:  number,  subtraction,  multiplication,  division,  and  the  extraction 
Do  not  change  of  roots  ;  and  it  is  plain,  that  the  nature  of  a  quan- 
;he  nature  of    ^.      j^   ^^^  ^^  ^^^  changed  by  prefixing  to  it   the  sign 

I'le  quantitiee.  •'  o  .-     a  ^^ 

which  indicates  either  of  these  operations.  The  quan- 
tity denoted  by  the  letter  n,  for  example,  is  the  same, 
in  everi/  respect,  whatever  sign  may  be  prefixed  to  it  ; 
that  is,  whether  it  be  added  to  another  quantity,  sub- 
tracted from  it,  whether  multiplied  or  divided  by  any 
number,  or  whether  we  extract  the  square  or  cube  or 
Ai  ebraic  ^ny  Other  root  of  it.  The  algebraic  signs,  therefore, 
"pns:         j^jjjj^    j^y   regarded    merely   as    indicating    opert./lo)ts    to 

how  regarded.  _ 

be    performed    on    quantity,    and    not    as    affecting   the 

nature  of  the  quantities  to  which  they  may  be  prefixed. 

Plus  and       ^,^[e   say,  indeed,  that   quantities    are    plus    and    minus, 

'  '°"''       but    this    is    an    abbreviated    language    to  express    that 

they  are  to  be  added  or  subtracted. 

I'rinciples  of  |2_    j^    Algebra,  as    in    Arithmetic    and    Geometry 

tll6    8ClGnC€  * 

FromwhM  all  tlie  principles  of  the  science  are  deduced  from  tnf 
deduced        ^j^fi,,iti(,„s   and  axioms;    and  the   rules   for   performing 

the  operations  are  but  directions  framed  in  conformifj 
Fxample        to    such    principles.       Having,    for    example,    fixed    b; 

definition,  the  power  of  the  minus   sign,  viz.,  that   an; 


INTRODUCTION.  11 

(jiiamily  before   which   it  is  written,  shall  be  regarded 

a^  to  be  subtracted  from  another  quantity,  we  wish  to  whatwewak 

to  discover. 

discover  the  process  of  performing  that  subtraction,  so 
a.-  ro  deduce  therefrom  a  general  formula,  from  which 
V.  c  can  fram(;  a  rule  applicable  to  all  similar  cases. 


SUBTRACTION. 

13.    Let    it   be    required,   for    example,   to   subtract    Subtractiom, 

from  h  the  difference  between  a  and  c.     \     . 

I      b 

Now,    having   written    the    letters,   with     j  Process. 

'  ^  \     a  —  c 

their  proper  signs,  the  language  of  Al-     '  - 

gebra  expresses   that   it   is   the   difference  only  between 

(/   and  c,  which    is   to  be  taken   from  h  ;  and  if  this  dif-      Diirerenc«. 

ference  were  known,  we  could  make  the  subtraction  at 

once.     But  the  nature  and   generality  of  the  algebraic 

symbols,  enable   us   to   indicute  operations,   merely,  and      Operations 

indicated. 

we  cannot  in  general  make  reductions  until  we  come 
to  the  final  result.  In  what  genei-al  way,  therefore, 
can  we  indicate  the  true  difference  ? 

If   we    indicate    the   subtraction  of  a 
from   h,  we   have   b  —  a.;    but  then  we 
have  taken   away  too  much  from  b  by 
the  number  of  units  in  c;for  it  was  not  a,  but  the  dif- 
ference   between   a    and   c   that    was    to    be    subtracted 
from  b.     Having  taken  away  too  much,  the  i-emainder 
1;    too   entail  by  c  :   hence,  if  c  be  added,  the   true   re- 
mainder  will   be  expressed   by  b  —  a  -\-  c. 


b  —  a 

Final  formula 
b  —  a  -{•  C 


Now,  by  analyzing  this  result,  we   see  that  the  sign      Analysis  l 

the  result. 

of  every   term   of  the   subtrahend    has    been   changed ; 
and  what   has  been  si  own  with   respect  to  these  quan- 


12 


INTRODUCTION. 


Oeneraliza- 
tion. 


Slil« 


titles  IS  equally  true  of  all  others  standing  in  the  ?ame 
relation :  hence,  we  have  the  following  general  rule 
for  the  subtraction  of  algebraic  quantities  ; 

Change  the  sign  of  every  term  tf  the  stibtrahendy  or 
conceive  it  to  be  changed,  and  then  vnite  the  quantities 
as  in  addition. 


MULTIPLICATION. 

Mnltiphca-  |^^  j^gj  ^g  j^^^  consider  the  case  of  raultiplicatioTi, 

tion. 

and  let  it  be  required   to   multiply  a  —  b  by  c.     Th*» 
algebraic   language    expresses    that    the 
Signification    difference    between    a   and    b    is    to   be 

of  ths 

taken    as    many     times    as    there    are 


language. 


Process. 


a  -b 
c 
ac  —  be 


lt>  nature. 


Priaeipls  for 
tk*  signs. 


units  in  c.  If  we  knew  this  differ- 
ence, we  could  at  once  perform  the  multiplication. 
But  by  what  general  process  is  it  to  be  performed 
without  finding  that  difference  ?  If  we  fake  a,  c  times, 
the  product  will  be  ac ;  but  as  it  was  only  the  differ- 
ence between  a  and  b,  that  was  to  be  multiplied  by  i\ 
this  product  ac  will  be  too  great  by  b  taken  c  times ; 
that  is,  the  true  product  will  be  expressed  by  ac  —  be-. 
hence,  we  see,  that, 

If  a  quantity  having  a  plus  Sign  be  multiplied  by 
another  quantity  having  also  a  plus  sign^  the  sign  of 
the  product  will  be  plus  ;  and  if  a  quantity  having  a 
minus  sign  be  multiplied  by  a  quantity  having  a  plus 
sign,   the  sign   of  the  product  will  be  minus. 


6*aeta,\caM»:         15.   Let  us^  now  take  the  most   general  ca.se,  viz., 
that  in  which  it  is  required  to  multipy  a  —  b  by  c  -  d. 


INTRODUCTION. 


13 


a 

~h 

c 

-d 

ac 

-be 

—  ad  -\-  hd 

ac 

—  be  —  ad  -\-  bd 

\\M  form. 


Lei  US  again  observe  that  the  algebraic  language 
'lenotes  that  a  —  h  is  to  be  ta- 
ker as  many  times  as  there 
are  units  in  c  —  d ;  and  if  these 
two  differences  were  known, 
their  product  would  at  once 
form    the    product   required. 

First :   let   us   take   a  —  b  as  First  step. 

many  times  as  there  are  units  in  c  ;  this  product, 
from  what  has  already  been  shown,  is  equal  to  ac  —  be. 
But  since  the  multiplier  is  not  c,  but  c  —  d,  it  follows 
that  this  product  is  too  large  by  a  —  b  taken  d  times  ; 
that  is,  by  ad  —  bd  \  hence,  the  first  product  dimin-  Second  step: 
ished  by  this  last,  will  give  the  true  product.  But,  by 
the  rule  for  subtraction,  this  difference  is  found  by  How  tuken. 
changing  the  signs  of  the  subtrahend,  and  then  uniting 
all  the  terms  as  in  addition  :  hence,  the  true  product 
is  expressed  by  ac  —  be  —  ud  -{-  bd. 

By  analyzing  this   result,  and  employing  an   abbre-     Analysis  oi 

.  xho  result. 

viated  language,  we  have  the  following  general  prin- 
ciple to  which  the  signs  conform  in  multiplication,  viz. : 

Plus  mulliplied  by  plus  gives  plus   in   the  product ;       General 

VrincipU. 

plus  multiplied  by  minus  gives  minus;  minus  mul- 
tiplied by  plus  gives  mimis ;  and  minus  multiplied  by 
minus  gives  plus   in    the  product. 

16.    The  remark   is  often   made  by  pupils  that  the       R«maA. 
above   reasoning  appears   very   satisfactory  so   long   as 
the   quantities    are   presented    under   the    above    form ; 

.        Particnlar 

but   why    will   —  !>   mulliplied    by   —   -    give    plus    'id  *  ^a^e 


14  INTKUDUCTION. 

How  can  the  product  of  two  negative  quantities  stand- 

mg  alone  be  plus  ? 
MiauBsign  In   the  first  place,  the  minus  sign  being  prefixed   to 

/  and  d,  shows   that   in   an  algvhraic  sense  they  do  not 

ta  interpre-     ^'tand  by  themselves,  but  are  connected  with  other  quan- 

'°°         tities  ;    and    if  they    are    not    so   connected,    the    minus 

sign  makes  no  difference  ;  for,  it  in  no  case  affects  the 

quaatity,  but  merely  points  out  a  connection  with  other 

quantities.       Besides,    the    product    determined    a^ove. 

being    independent   of  any   particular    value    attributed 

Foim  jf  the    ^^  ^he  letters  a,  b,  c,  and  d,  must  be  of  such  a  form  as 

product :        ^^  ^g  ^^.^^^  j-^^.  j^jj   y^lues  ;   and   hence  for   the   case  in 

must  be  irue 

forquantiiiei    which  «  and  c  are  each  equal   to  zero.      Making  this 

of  any  value 

supposition,   the  product   reduces  to  the  form  of   4-  bJ. 
Signs  in        The  rulcs  for  the  signs  in  division  are  readily  deduced 

division.  ,  -,    „    .  .  „,...  ,.  ...  , 

trom   the   definition   ot   division,  and  the  principles   al- 
ready laid   down. 

ZERO     AND      INFINITY. 

Zero  and  17.    The  tcrms  zero  and  infinity  have  given  rise  to 

"''y-  much  discussion,  and  been  regarded  as  presenting  diffi- 
culties not  easily  removed.  It  may  not  be  easy  to 
frame  a  form  of  language  that  shall  convey  to  a  mind, 

Ideas  not  but  little  versed  in  mathematical  science,  the  preci.>-e 
ideas  which  these  terms  are  designed  to  express  ;  bu( 
we  are  unwilling  to  suppose  that  the  ideas  themselvei 
are  beyond  the  grasp  of  an  ordinary  intellect.  The 
terms  are  used  to  designate  the  two  limils  of  Space 
and  Number. 

18.  Assuming  any  two  points  in  space,  and  joining 


INTRODUCTION.  15 

them  by  a  straight  line,  the  distance  between  the  points 
will  be  truly  indicated  by  the  length  of  this  line,  and 
this  length  may  be  expressed  numerically  by  the  num- 
ber of  times  which  the  line  contains  a  known  unit.  If 
nDw,  the  points  are  made  to  approach  each  other,  the     ii'.usttation, 

showing  lh« 

length  of  the  line  will  diminish  as  the  points  come  meaning  of 
nearer  and  nearer  together,  until  at  length,  when  the  ^  *™ 
two  points  become  one,  the  length  of  the  line  will 
disappear,  having  attained  its  limit,  which  is  called 
^ero.  If,  on  the  contrary,  the  points  recede  from  each 
other,  the  length  of  the  line  joining  them  will  con- 
tinually   increase ;    but    so    long    as    the    length   of  the    lUustraiicn, 

.  showing  the 

line   can   be   expressed   in    terms   of  a   known    unit  ot      me.nmgof 

.     f    • ,  -r»    i      -c  ii  the  term 

measure,   it    is    not    infinite.     But,   if   we    suppose    the       ,„fi„j,j. 
points    removed,    so   that   any   known    unit  of   measure 
would   occupy  no   appreciable  portion   of  the   line,   then 
the   length   of  the   line   is    said   to   be  Infinite. 

19.  Assuming  one  as  the  unit  of  number,  and  ad- 
mitting the  self-evident  truth  that  it  may  be  increased 
or  diminished,  we  shall  have  no  difficulty  in  under- 
standing   the    import    of    the    terms    zero    and   infinity.     The  term* 

Zero  and  In- 

as   applied   to   number.     For,   if  we   suppose    the    unit     fmity  applied 

,,,...,,,         ,.    .   .  ,1  to  numberg. 

one  to  be  continually  diminished,  by  division  or  other- 
wise, the  fractional  units  thus  arising  will  be  less  and  niustratioa. 
less,  and  in  proportion  as  we  continue  the  divisions, 
they  will  continue  to  diminish.  Now,  the  limit  or 
boundary  to  which  these  very  small  fractions  approach, 
is  called  Zero,  or  notliing.  So  long  as  the  fractional  Zwo : 
number  forms  an  appreciable  part  of  one,  it  is  not 
zero,  but  a  finite  fiaction  :   and   the  term   zero  is  only 


^.^ 


16 


tUnstr&tion. 


Infinity ; 


The  terms, 

how 
employed. 


Xtu  limits. 


Why  limits? 


INTIIODUCTION. 


Definition  of 
a  limit. 

Of  Space   and 
Number. 


Subject  of 

equatious  : 

how  divided. 

First  part: 


applicable  to  that  which  forms  no  appreciable  part  of 
the  standard. 

If,  on  the  other  liand,  we  suppose  a  number  to  be 
continually  increased,  the  relation  of  this  number  to  the 
unit  will  be  constantly  changing.  So  long  as  the  num- 
ber can  be  expressed  in  terms  of  the  unit  one,  it  ia 
finite,  and  not  infinite  ;  but  when  the  unit  one  forms 
no  appreciable  part  of  the  number,  the  term  infinite 
is  used  to  express  that  state  of  value,  or  rather,  that 
limit  of  value. 

20.  The  terras  zero  and  infinity  are  therefore  em- 
ployed to  designate  the  limits  to  which  decreasing  and 
increasing  quantities  may  be  made  to  approach  nearer 
than  any  assignable  quantity;  but  these  limits  cannot 
be  compared,  in  respect  to  magnitude,  with  any  known 
standard,  so  as  to  give  a  finite  ratio. 

21.  It  may,  perhaps,  appear  somewhat  paradoxical, 
that  zero  and  infinity  should  be  defined  as  '•  the  limits 
of  number  and  space"  when  they  are  in  themselves 
not  measurable.  But  a  limit  is  that  "  which  sets  bounds 
to,  or  circumscribes  ;"  and  as  all  finite  space  and  finite 
number  (and  such  only  are  implied  by  the  terms  Space 
and  Number),  are  contained  between  zero  and  infinity, 
we  employ  these  terms  to  designate  the  limits  of  Num- 
ber  and  Space. 

OP     THE     EQUATION. 

22.  The  subject  of  equations  is  divided  into  two 
parts.  The  first,  consists  in  finding  the  equation  ;  that 
is,  in  the  process   of  expressing   the  relations  existing' 


INTRODUCTION. 


17 


Solution 

Discussion  ol 
an  equation 


Statement : 
what  it  ia. 


between    the    quantities    considered,    by    means    of   the 
algebraic    symbols    and    formula.       This    is    called    the 
Statement   of  the    proposition.     The    second    is    purely      s-atomsor 
deductive,   and   consists,  in  Algebra,   in   what   is   called     "  ""   ^^ 
the   solution  of  the  equation,  or  finding  the  value   of 
the  unknown  quantity ;   and  in  the  other  branches  of 
analysis,  it  consists    in    the  discussion  of  the  equation  ; 
that   is,  in   the   drawing  out   from    the  equation   every 
proposition  which  it  is  capable  of  expressing. 

23.  Making  the  statement,  or  finding  the  equation, 
is  merely  analyzing  the  problem,  and  expressing  its 
elements  and  their  relations  in  the  language  of  analy- 
sis. It  is,  in  truth,  collating  the  facts,  noting  their 
bearing  and  connection,  and  inferring  some  general 
law  or  principle  which  leads  to  the  formation  of  an 
equation. 

The    condition    of  equality    between    two    quantities      ^^quaiity  of 

111.  „  tw  quanti- 

IS  expressed   by  the  sign  of  equality,  which   is   placed  ties: 

between    them.     The  quantity  on   the   left  of  the   si<^n       ^°^  ^"^ 

'^  pressed. 

of  equality   is    called    the    first    member,    and    that    on     i>t  .nembe, 
the  right,  the  second  member  of  the  equation.     Hence,     -''  "lember 
an   equation    is    merely  a    proposition    expressed    aWe-     I'ropositioo 
braically,  in   which  equahty  is   predicated  of  one  quan- 
tity as  compared  with  another      It  is  the  ffreat  formula 
!jf  Alsrebra. 


24.  Every  quantity  is  either  abstract  or  concrete  : 
hence,  an  equation  which  is  a  general  formula  for 
expressing  equality,  must  be  either  abstract  or  con- 
crete. 

2 


Abstract. 
Zoncteie. 


18 


INTRODUCTION. 


Abstract 
•qnation. 


Concrete 
equation. 


An  abstract  equation  expresses  merely  the  relation 
of  equality  between   two   abstract  quantities  :  thus, 

a  -^  b  =:  X, 

is  an  abstract  equation,  if  no  unit  of  value  be  assigned 
to  either  member ;  for,  until  that  be  done  the  abstract 
unit  one  is  understood,  and  the  formula  merely  ex- 
presses that  the  sum  of  a  and  b  is  equal  to  x,  and  is 
true,  equally,  of  all  quantities. 

But  if  we  assign  a  concrete  unit  of  value,  that  is, 
say  that  a  and  b  shall  each  denote,  so  many  pounds 
weight,  or  so  many  feet  or  yards  of  length,  x  will  be 
of  the  same  denomination,  and  the  equation  will  be- 
come concrete  or  clcnominate. 


Five  opera- 
tions may  be 
{wrformed. 


Axioms. 
First. 


Second. 


Third 


25.  We  have  .-een  that  there  are  five  operation? 
which  may  be  performed  on  an  algebraic  quantity 
(Art.  3).  We  assume,  as  an  axiom,  that  if  the  same 
operation,  under  either  of  these  processes,  be  performed 
on  both  members  of  an  equation,  the  equality  of  the 
members  will  not  be  changed.  Hence,  we  have  the 
five   following 

AXIOMS. 

1.  If  equal  quantities  be  added  to  both  members 
of  an  equation,  the  equality  of  the  members  will  not 
be   destroyed. 

2.  If  equal  quantities  be  subtracted  from  both  mem- 
bers of  an  equation,  the  equality  will  not  be  destroyed. 

3.  If  both  members  of  an  equation  be  multiplied  by 
the   same   number,  the  equality  will   not  be  destrojed 


INIRODUCTION.  10 

4.  If  both  members  of  an  equation  be  divided  by  Fourth, 
the  same  number,  the  equahty  will  not  be  destroyed. 

5.  If  both  members  of  an  equation  be  raised  to  Fifth. 
i:he  same  power,  the  equality  of  the  members  will 

not  be  destroyed. 

G.   If  the  same  root  of  both  members  of  an  equa-  sixth, 
tion  be  extracted,  the  equality  of  the  members  will 
not  be  destroyed. 

Every  operation    performed  on   an   equation  will      Use  of 
fall  under  one  or  other  of  these  axioms,  and  they 
afford  the  means  of  solving  all  equations  which  ad- 
mit of  solution. 

26.  Two   quantities   are   said  to  be  eqtial,  when  Equality  de- 
each  contains  the  same   unit   an   equal  number  of 

times.  Hence,  the  term  equal  applies  to  measures, 
and  has  the  same  signification  in  Arithmetic,  in  Equal  in  aii 
Algebra,  and  in  Geometry.  If,  in  Geometry,  two 
figures  can  be  applied  to  each  other,  so  as  to  coin- 
cide or  fill  the  same  space,  they  are  said  to  be 
equal  in  all  their  parts. 

27.  We  have  thus  pointed  out  some  of  the  marked 
characteristics  of  Algebra.      In  Algebra,  the  quan-    classes  of 

,.,.  i       1  •   1       n  •  •  quantities  in 

titles,  about  which  the  science  is  conversant,  are  Algebra, 
divided,  as  has  been  already  remarked,  into  known 
and  unknown,  and  the  connections  between  them, 
expressed  by  the  equation,  afford  the  means  of 
tracing  out  further  relations,  and  of  finding  the 
values  of  the  unknown  quantities  in  terms  of  the 
known. 


20  INTRODCCTION. 

SUGGESTIONS     FOR     THOSE     WHO     TEACH     ALGEBRA. 

re*s  rabji^  ^'  ^^  cart'ful  to  explain  that  the  letters  employed, 
are  the  mere  symbols  of  quantity.  That  of  and  in  them- 
selves, they  have  no  meaning  or  signification  whatever, 
but  are  used  merely  as  the  signs  or  representatives 
of  such  quantities  as  they  may  be  employed  to  denote. 
Signij  indicate        2.   Be  careful    to  explain    that    the   signs   which    are 

operations. 

used  are  employed  merely  for  tlie  purpose  of  indicating 
the  five  operations  which  may  be  performed  on  quan- 
tity ;  and  that  they  indicate  operations  merely,  without 
at  all  affecting  the  natui-e  of  the  quantities  before  which 
they  are  placed. 
Letters  and         g^    Explain    that   the  letters    and   signs    are  the  ele- 

si«rii«  elements 

of  language,    meuts  of  the  algebraic  language,  and  that  the  language 

itself  arises  from  the  combination  of  these  elements. 
Algebraic  ^    Explain  that  the  finding  of  an  algebraic  formula 

formula  ^  ->  ft 

is   but   the  translation   of  certain   ideas,  first   expressed 

in  our  common  language,  into  the  language  of  Algebra; 

[«■  interpreta-    and   that  the  interpretation  of  an  algebraic  formula   is 

tion. 

merely  translating  its  various  significations  into  common- 
language. 

Tjanguage.  5.  Let  the  language  of  Algebra  be  carefully  studied, 

so  that  its  construction  and  significations  may  be  clearly 
apprehended. 
Co-efficient.         6-   Let   the   difference  between  a  co-efficient  and   an 

E«ponent  cxpoueut  be  carcfully  noted,  and  the  office  of  each  often 
explained  ;  and  illustrate  frequently  the  signification  of 
the  language  by  attributing  numerical  values  to  letters 
in  various   algebraic  expressions. 

7.  Point  out  often  the  characteristics  of  similar  and 


INTKODUCTION.  0J 

iissiniilar  quantities,  and  explain  which  may  be  incor-        simiiax 

^uaatiti** 

porated  and  which  cannot. 

8.  Explain  the  power  of  the  minus  sign,  as  shown     Minus  sign 
in  the  four  ground   rules,  but  very  particularly  as  it  is 
illustrated  in  subtraction  and  multiplication. 

9.  Point  out    and    illustrate    the   correspondence   be-     Arithmetic 

and  Algebra 

(ween   the  four   ground   rules  of  Arithmetic  and  Alge-      compared. 
bra  ;  and  impress  the  fact,  that  their  differences,  where- 
ever    they    appear,    arise    merely   fi'om    differences   in 
notation    and    language  :    the    principles    which    govern 
the  operations  being  the  same  in  both. 

10.  Explain  with  great  minuteness  and  particularity.      Equation. 
all    the   characteristic    properties  of  the   equation ;    the    ^'^  propertjea 
manner  of  forming   it ;   the  different  kinds  of  quantity 

which    enter   into  its   composition  ;    its  examination   or 
di-cussion;  and  the  different  methods  of  elimination. 

11.  In  the  equation  of  the  second  degree,  be  careful    Equation  of 

the  second 

to  dwell  on  the  four  forms  which  embrace  all  the  cases,        degree. 

and    illustrate   by  many  examples    that  every  equation 

of  the  second  degree  may  be'  reduced  to  one  or  other 

of   them.      Explain    very    particularly   the   meaning  of       its  form. 

the  term  root;  and  then  show,  why  every  equation  of       itsroou. 

the   first  degree    has    one,   and    every   equation   of  the 

second  degree  two.     Dwell  on   the  properties  of  these 

roots  in  the  equation  of  the  second  degree.     Show  why 

their  sum,  in  all  the  forms,  is  equal   to  the  co-efficient     Thtirsnm. 

of  the   second   term,  taken   with  a  contrary  sign  ;   and 

why  their  product  is  equal  to  the  absolute  term  with  a  Their  prodnet 

i-ontrary  sign.      Explain   when   and   why  the   roots   are 

iniairinarv. 


22  INlRObUCTlON. 

General  12.   In  fine,  remember  that  every  operation  and  rule 

nncip  e^       ._^  based  on  a  principle  of  science,  and   that  an  intelli- 
gible  reason    may  be  given  for  it.     Find   that  reason, 
and  impress  it  on  the  mind  of  your  pupil  in   plain  and 
Shoaid  be      simple  language,  and  by  familiar  and  appropriate  illus- 
trations.     lou  wdl  thus  impress  right  habits  ot  invei?- 
tigation    and    study,   and    he    will    grow   in    knowledge. 
The    broad    field    of    analytical    investigation    will    be 
opened    to    his    intellectual    vision,    and    he    will    have 
made  the  first  steps  in  that  sublime  science  which  dis- 
They  lead  to    covcrs  tlic  Uiws  of  nature  in   their  most  secret  hiding- 
««nerai  laws,    pjac^s,  and  follows  them,  as  they  reach  out,  in  omnipo 
tent   power,  to  control   the   motions   of  matter   through 
the    entire    regions    of  occupied    space. 

(See    Davies'  Nature  and    Utility   of    Mathematics 
Article  Algebra). 


KEY. 


1. 
2. 


NUMERICAL      VALUES. 
Art.  16,  p.  22. 

12x3+3x4  —  5x2  —  3x1  =  36.     Ans. 


5x4 


2 


+ 


2+3       3— 1' 2+2x1 

.       A       o      3x3x1 
4x4x2 


=  5.     Ans. 


8x4 


4. 


2+1  5x3+1 

6  X  13       16  X  22       12  X  32 


=  27.  Ans. 


+ 


+  4  =  10-^.  Ans. 


32       4    '5x1* 

5.  (4  +  9  +  16  +  25)  X  (9  +  16)  =  54  x  25  =:  1350.  Ans. 

6.  4  +  25  +  16  +  25  X  (9  +  16)  =  45  +  25  x  25 

=  670.  Ans. 

7.  4  +  (9  +  16  +  25)  X  (9  +  16)  =  4  +  50  x  25 

=  1254.  Ans. 

1  +  4  +  9   9-1   14   8   ^,   . 

^-  --r.:rT-^2^^  =  y^4  =  ^'^'  ^^^■• 

9.  (1  +2)2  X  (1  -2  X  1  +4)  +  (1  +  3)2  X  (1-1x3  +  9) 

=  139.  Ans. 


10.  1  X  (4  +  9)  +  4  X  (1  +  9)  +  9  X  (1  +  4)  =  98.  Ans. 


24  KEY    TO    UAVIES'    BOUKDON.  [19-26. 


ADDITION. 

Art.  19,  p.  25. 

1.  (4  +  11)  _  (8  +  9)  =  -  2.     .-.  -  2a2J.    Ans. 

2.  (7  +  6)-(l  +  7  +  8)  =  -3.     .'.  -dabcl     Ans. 

3.  (9  +  15  -  24)  cb^  +  (9  -  8)  ac^  +  Sea  =  a(^  +  Sca.  Ans. 

4.  (6  +  7  -  13)  «c2  +  ( _  5  _  3  _^  18)  ^js  ^  io«^<^     Ans. 

5.  (1  —  9)  abc^  +  (—  1  +  0)  abc  +  (5  —  8)  ac^ 

=  —  Sabc^  +  5abc  —  3ac^.     Ans. 

6.  (3  -  9)  a^^  +  (_  7  +  9)  a^b  +  (5  +  3)  ab 

=  _  6aW  +  '2(fib  +  Sab.     Ans. 

7.  (3  -  3  -  6)  ac¥  +  (_  7  +  6)  a^cW  +  (_6  +  4  +  2)  a^h^ 

=  —  a^cW  —  Q)ac¥.     Ans. 

8.  ( -  7  +  6  +  1)  aWc^  +  (9  —  5)  a-bc^  -  b^c^ 

—  ^a^b(?  —  b^(P.    Ans. 

9.  (-10  + 7-5)a3j  + (6  —  5 +  3)«2j2 

= .  —  Sa^  +  4a2^2.    Ans. 


Art.  20,  pp.  26-29. 

e.  da+    b 

3a  +  3b 
—  9a  —  76 
6a  +  9b 
8a  +  3b  +  Be 

11a  +  96  +  8c.  ^ws. 


26-27.]  ADDITION.  25 

7.  Sax  +    3ffc  +     /  ; 

—  9ax  +  Ha  -\-  d. 
6ax  +    Sac  +    3/ 

Sax  +  ISac  +    9/ 

Sax —  14/^ 

llax  +  19a6'  —     /  +  7a  +  6?.     ^W5. 

8.  Sa^c  +  5a5 

Vrt^c  —  Sab  +    3«c 
5a^c  —  Gab  +    9«c 

—  8»^c  +    ab  —  I2ac  - 
larc  —  Sab.  Ans. 

9.  lMh%  —  12ah'b 

ha'^x^  +  Ibahb  —  lOax 

—  %a^x?b  —  ISa^cb 

—  \Sa^x%  —  Ua^eb  +    9ax 

4:a^a^b  —  22a^cb  —      ax.    Ans. 

10.  Sa  +    b  -^    c 

•  6a  +  '-iZ*  +  Sac 

a  +    6'  +    ac 

—  Sa  —  Sb  —  Qac 

6a  —  5b  -{-  2c—  5ac.    Ans. 

11.  ba^  +    Gcx  +  9Jc2 

—  Sa^^b  +    7ca; 

2a2^,  _  i^cx  —  %<? 

—  a%  —    2cx.  Ans. 

12.  Sax  +    bab  +  3a2JV 

—  ISax  +  10«5  +  6a2 

IQqjg  —  15aZ>  —  GaWc^ 

—  SaWc^+Qa\     Ans. 


26  KEY    TO    DAVILS'    BOURDON.  [27. 

13.  Ala^b'^c  —  27abc  —  Ua^y 

lOa^o^c  +    9abc 

6la^lPc  —  ISabc  —  14a2«/.  Ans, 

14.  ISabc  —  dab  +  Gc^  —  3c  +  9aa; 

9abc  +  3c  —  9a.r 

27aic  —  Oab  +  Gc^.  ^ms. 

15.  Sabc  +      b^a  —  2ca;  —  ^xy 

—  \Wa  4-  7-T  —    yy 

8a6c  —  13i%  +  hcjc  —  Ixy.    Ans. 

16.  9a2c  —  14f<%  4-  15«2^,2 


17. 


8a2c- 

Uaby  +    7am.    . 

I'^aW  + 

%a%  -  3^2 

—  14a5J2 

+  7^2  _  9^3 

Va^^^  — 

15^3,^             —    a^ 

+ 

14^3^ 

— 

lOa^Z* 

18.  3«2;2  —    7aa;3  —  llaxy 
9aa;2  +  ISaa;^  _|_  ^^axy 

—  lax^  +    3aa^  +  7as&  +  ^^ga; 

baoi?  +  14aa;3  _|_  17^x2/  +  la^b  +  4:Z>c.c.    ^ws. 

19.  3a2+    bd^m^—    daH 
7a2  _    8a-'^'2c2  —  lOa^a; 

+  16aVj3c2  +  19a3a;  +  Ipg^ 

10a2  +  13a'iV  +  lOaJ.     Am. 


27-30.]  SUBTRACTION-.  27 

20.  '^a^  —  3nhc  —  8b\- —    9(^  +    cd^ 

—  5cM  +  8abo  —  W^c  +    Sc^  +    ccP 

^a^b -f  %'ic  —    Srg —  3# 

31.  —  18a3^  +  2a¥  +    6a262 

7a3^-8a^/4-    haW 

—  5a%  +  6r/&^  +  llaW 

33.  3a3*2c  —  lGa*.i:  —    OfAa;^^^ 

_  SaWc  —      g-^.^;  +  Ukuhl 

aWc  +      «a:^r7.     ^?^5. 

23.  8«5  4-  (—  10  —  13)a^5  +  (_  16  +  15  -  16) a^^-'  + 
(4  4.  24  +  30)  a%^  +  (—6  +  33)  ai^  —  m  =  8a^  —  22a*b 
—  \laW  +  i8am  +  36flA4  —  8^»5.     ^^^.s. 

34.  a^  —  f  «2x-  +    \ax^  -   ^ofi 
a^  +  \a^x  +    ^«a;2  +    fa^s 

3a3  —  ig^^;  +  ^^a.r2  +  ^^ofi 

4a2  —  |«2^  ^  ||-«.f2  +  ^igx^.     ^ws. 

SUBTRACTION. 

Art.  22,  pjt.  30-31. 

6.  3a2a;  _  iSabc  +  7a^ 
9a^x  —  ISabc 

—  6a^x  +  7a2.     Jw5. 

7.  bla^'^c  —  ISnbc  —  Ua^y 
Ala^^c  —  27abc  —  Ua^y 
\QaWc  +  9abc.  Ans. 


28  KEY    TO    DA  VIES'    BOURDON.  [30. 

8.  ^labc  —  9a5  +  Gc^ 

'dabc  +  3c  —  %ax 

ISabc  —  9ab  +  Gc^  —  3c  +  9ax.    Ans. 

9,  SflJc  —  1 25^a  +  bcx  —  '\xy 

—  135%  -f  '^cx  —    xif 


10. 


SaJc  + 

¥a  —  '%cx  —  %x]).    A 

8fl2c- 

9a2c- 

-  Uaby  +    naW 

-  Uabt/  +  15aW 

—    c?c 

—    Sa^'i     A)is. 

11.  9rt6a;2  —  13  +  20ab^T  —  4:¥cx^ 
9a6a;2  —    6  +    'Sab^x  +  3^>6c.^^ 

—    7  +  17aJ3;j;  _  7J6(;^2_     4ns. 

12.  5«*  -    7«=5/>2  -  dr-(P  +  7f/ 

3a^  —  16a^^  —  IthP  —  3fl^ 

2a*  +    8^352  +  4.AP  +  7r/  +  3^2.     Ans. 

13.  51«2^<2  _  48^3^  _(_  10^4 
_    6ft2^^  —    ^a%  +  lOff* 

57^2^2  _  40fl35.  ^ws. 

14.  21^y  +  2bxhf  +  eSz?/-'  —  40^5 

+  Ux^y^  +  48y^4  —  40z/5 
21a^?/2  —  39x2/  ^  30a:?/*.  ^ws. 

15 .  53x3?/2  _  l^:^iy^  _  1  s.r*?/  —  5(  a;' 
18a:3y2  _  i5a,-2y3  ^  24.r*y 

35a;3/  _  ^^x^y  —  56a;5     ^^g. 


31-32.]  SUBTKACTION-.  29 

16.  lOa™  —  Ub^  —    cV  -\-    5^1 
—    9a™  +    2^)°  +    cP  —    ocZq 

19am  _  i7jn  _  26-P  -f  lOcK     Ans. 

17.  1^  +    36  +  ic  +  i*c 
|a  —    76  +  y  —  Uc 

—    rt  +  112  —  ir  ^- 1^><\     ^ws. 

18.  i^  +  (-|  +  i)^'-  !■'• 


^rf.  ;?5,  i>/>,  31-32. 

1.  3a;  —  7?/  —  a;  —  3^/  —  4?/  +  7a;  =  9a;  —  \^.     Ans. 

2.  «  +  ^  +  c  —  «  —  Z'  +  c  —  a-fZ»+c 

=  —  a  -\-  1)  +  3c.     ^ws. 

.  3.     a;2  —  23^2/  +  ?/'  —  a;^  —  xy  +  y^  +  2.ry  —  x~  —  y^ 

=  —  a-^  —  xy  +  y^.     Aus. 

4.       «2  _  J2  _^  ^  _  J2  _,_  C2  —  a2  +   C2  —  ^2  +  ^,2 

—   «2  _  3^,2  ^  3^2.       ^,i^.. 

6.  2a  —  [2a  —  Z*  —  2a;]  +  [&  —  2a;  +  2b] 

=  2a  —  2a  -\-  b  +  2x  +  b  —  2x  +  2b  =  4.b.     Ans. 

7.  [1  4-a._l  _2a;]  +  [i_.^+l_2a;]  —  [1— .t— l  +  2a;] 
=  l+a;  —  1  —  2a;  +  l—  a;  +  l  —  2a;  —  l+a;+l  —  2a;  = 
2  —  5.1'.     A  ns. 

8.  a2  ^  ^ab  +  b"^ —{a^  +  ab  —  t^  —  2ab  +  a^  +  J2] 

=  a2  +  2a^>  +  ^>2  _  a2  _  rtj  +  J2  _i_  g^/j  _  «2  _  ^: 

=  —  a2  +  3aS  +  b\     Ans. 


30  KEY    TO    DAVIES'    BCUr.DOX.  [32-38. 

9.     Ux  —  \7x—  [8x  —  Qx+Cx]] 

=  11:»  —  ^7a;  —  8x  +  9x  —  C).r\ 

=  llic  —  7a;  -f  8a;  —  9a;  +  6x  =  9x.     Ans. 

10.     4a;  —  3y  —  \-ix  +  4^  +  3a;  +  [y—9x—2ij-\-x  +  x—i/]\ 
=  4:X—'3i/—\2x-\-4:y  +  Sx-\-y  —  dx—2y  +  x  +  .i—//^ 
=  4:X—oy—2x—4:y—3x—y  +  9x-^2y—x~.r+)/ 
=:  6a;  —  oy.     Ans, 

Note. — Many  of  the  above  examples  can  be  abbreviated  by  reduction. 

MULTIPLICATION. 
Art.  28,  p.  :i5. 

16.  -  2  X  3  X  4a2^i+i  Ja;i+2+iyi-iM  -  _  24:a^bxy.    Ans. 

17.  +  3  X  2  X  3Ja;y+i0i+3  =  ISb.ryz^     Ans. 

18.  +  3/w««i+i+2_gP^p+q  _  S-irina^xVyP'fi.     Ans. 
Note. — The  student  should  add  the  exponents  mentally. 

19.  —  ^   X   ^   X  3    X  2a2cP+q+3^m+q+l2 

20.  —  20  +  I  X  laP^'^x^+P+^y  =  — «P+ia;P+q+i«/.     Aiis. 
31.     —  I  X  I  X  ix^y^  =  —  -^iX^y^.    Ans. 

22.     +  i  X  i  X  i  X  ix*y^  =  tV-^V-    ^^s- 

Art.  SO,  p2).  38-39. 

12.  2a2—    3aa;  +    4a;2 

5a^  —    6aa;  —    2x^ 

10a*  —  lort^x  +  20a2a;2 

—  12a3a;  +  18a2a;2  _  24«.r3 

—    4a2a:2+    6aa;3  — 8a;^ 

lOrt*  —  2'7a^x  +  34cf2r2  _  igaa^  —  8x*.     Ans. 


38.]  MULTIPLICATION.  31 

13.  Sx^  —  ^x  +5 

a;2  -}-  2xy  —  3 

+  Ga;^^  —  4a;2y2  _j_  iOa;y 

-'^^ —  9a;3 +    6a:y  —  15 

3a:*  +  ^x^ij  —  4a.2  —  ^x'y'^  +  IGxy  —  15.     Ans. 

(14.) 
.3a:3_|_2a;y  +  3^2 

3a:3_3a;22^2.^5^3 


6a;6  +  4a;5/+6a;y 

—dx^y^  —ex*i/*—9T^y* 

+]5a:3?/3  +  10a:y+15yS 

6rc6_ 5a;y  +  6x^y^—  Gcc^y^—dxY  +  ISa:^?/^  +  lOa;^  + 1 5^5,  A ns. 

15.  8ffa;  —  Gab  —  c 
2ax  +    ab  -\-  c 

16a2.c2  —  12a2fe  _  2acx 

+    8a2^a;  —  6«2J2  _    „^(. 

+  Sarx  —  Gabc  —  c^ 

16«2a;2  —    4,a^x  +  6aca:  —  QaW  —  lobe  —  (?■.    Ans. 

16.  3a2  —  5*2  +  3c2 

ff2_      J3 


3a*  —  5a2J2  _|.  3^2^ 

—  3a2^>3  -I-  5^5  _  3^^c2 

3a*  —  50.2^,2  +   3^2^2  _  3^^2J3  ^_  5J5  _  3^^        ^f^g^ 

17.        3a2  +  5bd  +    c/ 

—  5«2  +  Abd  -  Sc/* 

—  15a*  +  25a^d  -    oa^cf 

+  12a^d                  -  20m^  +    Ucdf 
— J4a2/ 4-  40bcdf  -  Sc^p 

—  15«*  +  ^1a%d  —  29a''cf—20bW  +  Ubcdf  -  Sc^/a.  An,^. 


32  KEY    TO    DAVIES'    BOUEDON.  [39. 

18.         AaW  —  M~Wc  +  M'hc^  —  7>a^(?  —  lah(^ 

2a¥  —  ^abc  —  2hc^  ■\-  c^ 

-S  f  +  8«^Z>4  —  lOa^^^^  +  16a3JV2  —  QaW<fi  —  X^a^l^c^ 
I  I    _  l6a*Z»-%  +  2QaWc^  —  ^'MW<^  +  Ua^c*  +  28a2&V 

I  [  +  i,aW(?  —  haWd^  +  Sa^^c^  —  '6d^c^  —  lahc^ 
^  .  ^a^l)i  _  \Q>,rh\c  +  28a36V2  —  34f<3JV3  —  4ff2/v3f3 
I  \  —  Wa'b^c  -T  Ua^c"^  +  7aWc*  +  Ua^^c^ 


e  s 

r' 


Explanation. — Tlie  partial  products  are  not  arranged  so  as  to  bring 
similar  terms  in  the  same  column,  on  account  of  the  difficulty  of  getting 
them  on  the  page. 

19.  a;2  —  la;  +  i 


-yX       g-X        -f-  gX 


20. 


1 

+ 

a;2 
2a 

1 

"2a-'' 

+ 

0-3 

4a2^ 
4a2^ 

X 

+ 

2a 

— 

4a2^ 

2a  4a2  8a^ 

1  J^  1      , 


39.]  MULTIPLICATION.  3j3 

21.     a^  +  3^>°  —  2cP 


a2m  _|_  3am^n  _  2a«»cP 

_  ^a^}p.  —  9J2II  +    6J°cP 

+  2«mcP  4-    6J°cP  —  4c^P 

a^m  —  9i2n  4-  126°cP  —  4c^.     .'l  ;/.v. 

22.  a;-  1  a;2  —  1  (1)     .t2-1 

a;  +  1  x^  +  1  (2)     a:*— 1         

2;2  —  a;  x^  —  x^  a:fi—x*—x^-\-i.  Anff. 

-^  X  —  1  +a;2_i 

(1)     x^  171  ^  ^     (2) 

23.  a;2  +  a;  +  1 
a;2  —  a;  +  1 
iC*  +  a;^  +  a;2 

—  :c2  —  x^  —  X 

x^  +  X  ■\-  1 

x^  -\-  x^  -\-  1     .     .     .     First  product, 
x^  +  X  —  1 
a^  +  a:*  +  a:2 

+  a;^  +  ar*  -f-  a; 

—  a:^  —  a:^ — _1 

ofi  +  x^  -{-  x^  -\-  X  —  1.    Ans. 

24.  a:  4-  3 

a;  +  5 


a:3  +  3a: 

+  5a;  +  15 

a:^  -f  8a;  +  15     .     .     .     First  product. 

a-  +  7 

a?  +    8a;2  +  15a; 

+    7a;2  +  56a:  +  105 

a?  +  15a;2  4-  71a;  4-  105.    Ans. 


34  KETTO       DAVIES'    liOURDON.  [43-46. 

DIVISION. 

Art.  36,  pp.  43-44. 

1.     -t^a;2-i  =  2z.    Alls.         2.     -i^ff2-i;^y3-i  _  Qaxy^.    Ans. 

Note. — The  subtraction  of  exponents  is  made  mentally. 

3.     ^abx  =  labx.  Ans.        4.     ^^%(?  =  —  8a^c^.  Ans. 

5.  iA^a^i)2cd*  =  —  ^a^i^cdK     Ans. 

-36 

6.  ^^^(fbcx  =  Idabcx.    Ans. 

7.  =:^^,tbcx  =  —  lOabcx.    Ans. 

8.  zi^&hcx*  —  —  IGbcx*.     Ans. 

10.  ^bo^-^  =  —  hbc^~°.    Ans. 

—  5 

Art.  39,  p.  45. 

11.  ^ar^y-\    Ans.  14.     >-a^c_j^b+i.     Ans. 

12.  —  i^b'^x^.     Ans.  15.     Imx-'^y^.     Ans. 

13.  _  ^x-h-\      Ans.  16.      2.Tn^m-2n.      ^^5. 

^r«.  40,  />.  40. 

1.  Dividend.     6a^x*if  —  12a^7^i/^  +  Iba^a^y^  \  'ha^7?y^    Divisor. 
Qiwtient.      2x'^y^     —    4caxy*    +    ha^Q^y.     Ans. 

2.  l2aY  —  IQaY  +  20gy  _  28a'y^  \   —  4ay 
—3y^  +    Aay^  —    bd^y   +    7a^.     ^«s. 

3.  ISa^^r  —  2Q)acy'  +  5crf^  |   —  hahc 

4  <?^ 

—  3fl    +    ^b-hf—  a-^b-^d^^  =  -  3fl  +  ^y^ t-    Ans. 

4.  6a^'°  —  12fl3ma;m  ^_  3flrm^^2m    |    3flm-l^m 

2rt3m+i.>r«i  —  4a2nin  _|_  ax^.    Ans. 

5.  2^:^"  —  Sa-^C"-^^  +  ea:^^"^)  {   Sa;"^-* 
|a;m^2  _  a;m  _|_  2a;m  2_     ^^^., 

6.  —  g'^  —  |a'°^°-P  1    —  \a^ 

2a™  4-  3a°~P.    .4«s. 


49.] 


1. 


2. 


6. 


8. 


9. 


DIVISION. 


35 


lOab  +  15ac  |  5a 
2b    +    3c.     A71S. 
30ax  —  54a:  |  6x 
Ans. 


Art.  41,  pp.  49-51. 

3.  10a;2j/  —  15 f  —  hy  j  by 
2^    —    '6y  —  1.       ' 

4.  13a  +  3a^-  —  ISaa^ 


6a    —    9, 


i    + 

6aa;^  +  Qa^a;  +  a'^-^^  \  ax 

Ans. 


X 


—    6a:2. 


An8. 
3a 
A  ns. 


ex     +  9a 

a^  4-  ^aa;  +  a;' 

a^  +    aa; 

aa;  +  x^ 
ax  +  a;2 


+  a  X. 

•2       a  +  « 


a  4-  a;.     Ans. 


0 


7.  a3  _  3^2^  _|_  Say' 


'^  —  f 


a 


,2  _ 


a^y 


a  —  y 


—  2a^y  4-  3a^2 


a^  —  'lay  +  y\    Ans. 


af-  —  f 

af^2jl 
0 

2\a%  —  VZa^cifi  —  Qab  \  —  Qab 
— 4a  +  2a^cb  —  L    Ans. 

6ar*  —  96  \  3x  —  6 

6x*  —  12a^         ■  2x^  +  ^"^  -\-  Sx  +  16.    ^ns. 

12a;3  _  96 

12a;^  —  243:2 

24a:2  —  96 
24a:2  _  ^Sx 

48a:  —  96 

48a:  —  96 

0 


S6 


KEY    TO    DAVIES'    BOURDON". 


[sa 


(10.) 


a^—2a*x+     a^a? 

—Sa*x+   9a^x^—l0a^x^ 
—3a*x+   6a^x^—  Sa^a^ 

a^—'da^x  +  Zax^—a? 

3«3^2_  7a2^  +  5aa4 
3a3^2_  Ca2a;3  +  3a^ 

—     a^x^  +  2a'X*—x^ 

0 

11.     ^Sx^-7Qax^-64.a^x  +  105a^ 

2x—3a 

4:8^—72ax^ 

24:X^—2ax—35al     v 

—  4aa;2 — 64:a^x 

-70a2a;  +  105a3 
—  70a2x  +  105fl3 

Ans. 


0 


12.    f—3fx^  +  dfx^—a^ 
y^ — Sy^x  +3y*x^ — y^.7^ 


y^ — 3yh>  +  3yx^ — a? 


y^  +  3y^x-\-3yx^  +  x^.     Ans. 


Sy^x  —  6y*x^  +   y^x^  +  3y\i^ — a* 
3fx  —  Qyh;^  +  9y^a^ — 3y^x*' 

3yh^—8y^a^+6tfx^—ci^ 
3y  V — 9y^3^  +  9.^2x4  _  3ya;5 

y^.i^—3y^x*  4-  3yx?—x^ 
y^x? — 3y^x^  +  3?/a;^ — a:* 

0 

Note. — In  examples  like  this,  bring  down  all  the  terms  and  arrange  as 
before. 


50.] 

DIVISION. 

13. 

e4a*b^  —  -IbaW 

8aW  +  5ab* 
SaW  —  bab\ 

—  AOaW 

—  40a3§' 

-  25aW 

-  2haW 

m 


14. 


Ans. 


0 

6a3  +  23a25  +  22aJ2  _,_  553 

0 


2a  +  5  J.     Ans. 


15.        6aa:g  +  6a.r2z/6  +  42fl;V  |  aa;  +  5ax  =  6aa; 
a:^  +      a:  ^^  +    "lax.     Ans. 

(16.) 
-15««  +  S7a''bd-29a''cf-20b"-(P  +  W)cdf-Sc'^p 
-\U^  +  25a}bd-  M^cf 

\2a%d-2ia'^cf-2Qi¥d^  +  44ftc(// 
\%a?bd  -20bH-'+  ^cdf 

+  iObcdf-8c^r 
+  40bcdf-Sc^P 


3a?-nbd  +  cf 


-5a^  +  46d-8c/.    Am. 


-2ia'cf 
-2Wcf 


17. 


a:*  —  x^y  +  2;2?/2 
x^y  +y 


x^  —  xy  +  2/2 


a:^^  —  .f2?/2  _|_  xy^ 


^2y% 


0 


Ans. 


08  KEY    TO    DAVIES'    BOURDON.  [50. 

X  —  y    


18.  a^  —  ^ 

a:*  —  a^y 


a?  +  x^y  +  xy"^  +  y^.    Ans. 


a?y  —  f 
7?y  —  x^y^ 


xY 

-t 

xY 

-xy^ 

xif  - 

-t 

xy^- 

-t 

0 
19.    3«*  —  8a2J2  +  3a2c2  +  55*  —  SSV  \  a"  —  U^ 


I- 


3«^  —  3a^^>^ I  3a2— 5^  +  3cg.     ^ ns. 

—  baW  +  3aV  +  5J*     . 

—  baW  +  5Z»* 


.(?-•■  •" 


3flV  —  3^>V 

0 


20.  In  this  example,,  we  see  that  62-^  divided  by  Zoi?  gives  2'jfi, 
and  that  15_v^  divided  by  ^y'^  gives  5?/^,  and  these  terms  are,  from 
the  nature  of  the  case,  both  terms  of  the  (juotieut.  The  product 
of  the  divisor  3.^3  +  •2x^y'i  +  32/^  by  22-3  +  5//  is  6.t«  +  4^:5^2  ^  Q^yi  _^ 
i5a-^?/^+10.r2?/5  4-i5y5.  ^}jig  subtracted  from  the  dividend  gives 
for  a  remainder,  —  Qa:^^^ — 6a:*?/*—  9a;y .  Starting  with  this  result 
for  a  dividend  and  with  the  partial  quotient  already  found,  we 
finish  the  operation  as  shown  below.  This  method  often 
shortens  the  operation  when  the  polynomials  are  large. 

_  9y5y2  _  6.r4v*  —  ^x^  '^^+  ^f  —  3a%2.     Ans. 
0 


50-51.] 


21. 


DIVISION. 


31) 


22. 


23. 


24. 


25. 


8  +  12a:-i  _,_  2.^-2  +  2^:-*    4  —  2r-^  +    a;-'^ 

8—    4;?;-'  +  2x-^  2  +  4a:-i  +  2a;-=^.     ^w*-. 


16a;-i  +  0       +  2x-* 

16a;-»  —  8a;-2  +  ix'^ 

8:c-2  —  4.C-3  +  2ar* 

Sx~^  —  4a;-3  +  2x-* 

0 

a;2n  _  ^2n     j  a;a  4,  yn 

a-gn  ^  a;nyu  I  g-n  _  ^n       ^^^_ 


—  a;°?/°  —  i/2n 

—  a:°?/"  —  v^ 

0 
a;3n  _  Sx^y^  +  3a;°?/2n  —  p' 
^  —    x^y^ 

—  2x^y''  -f  3a;°2/2n 

—  2a;^"i!/°  +  2x'^y^ 

a;ny2n  —  ySn 

X°y^  —  yS"! 

0 

x^  — 'X>  —  x-^  -\-  ar* 
X*  —  x^ 

—  x-^  +  x-^ 

—  x-^  4-  X-* 

0 


r3n  I  a;n  _  yn 


a;' 


.2n 


—  2a;°«/°  +  y2n.  ^„,^. 


a; 


•ari 


a^  —  x~^.     Ans. 


a;m+i  _|_  x^y  +  .T?/°^  4-  2/™+i       a;™  +  ^™ 
2.m+i  _j_  xy^  X    +  y. 


+  y.     Ans. 


x^y 
x^y 


0 


40 


26. 


27. 


28, 


39. 


KEY    TO    DAVIE6'    BOL'RDOX. 

a°  — ^>° 


[51. 


2a^—2a^H'' 

_4(;t2D^n^_4aU//n 

2an62n_2J3n 

2a°62°— 2^o 
0 

1  +  X 


2a'^—4:aH''  +  2b^.    Ans. 


1 

1  +  a; 


1  —  X  +  x^  —  a?  +  xi^,  etc.    Ans. 


X 


—  X  —  x^ 


+  a;2 

+  x^  -\-  a? 


—  ^ 

—  X?  —  x^ 

x^ 

!  1  —  2a;  +  a;2 


1  _  2a;  +  2;2      i  ^  22;  +  3:r2  +  4a;3_  ^tc.    Ans. 


2x  —  x^ 

2x  —  4a;2  +  2a^ 


3a;2  —  2x^ 
3a;-  —  6a;3  +  3a;^ 
4a;3_3a4 


x  —  a 
x  —  b 


x  —  h 


\  —  {a  —  b)  xr^  —  b{a  —  b)  a;~2,  etc.    Ans. 


-{a-b) 

—  {a  —  b)+b{a  —  b)  a;-^ 

-b{a  —  b)  a;-i 


51.] 

DIVISION. 

30. 

^-^x^+    i^-  +  i 

x'i  —  2x-\-^ 

a^  —    2^3  _f.    |a;2 

x^  +  2^  +  I 

3a;3  —  ^x^  +  -i-a; 

2x^—    ^x^   r    X 

^x^  -^x  +  i 

^x^  —  fx 

+  i 

41 


^;is. 


0 


Note. — In  Ex.  31,  bring  down  the  entire  remainders  and  arrange  them 
with  respect  to  the  descending  powers  of  x,  and  the  ascending  powers  of 
y  and  z.  In  Ex.  82,  arrange  the  remainders  with  respect  to  the  descending 
powers  of  x  and  y,  and  the  ascending  powers  of  g,  as  far  as  possible. 


31.    x^-{- dx^y  +  3xy'^ -i-y^  +  z^\  x  +  y-\-z 
x^-\-  x^y-\-  xh 


x^-{-2xy-\-y^ — xz — yz  +  z^.  Ans. 


2x^y  +  Sxy"^  +  y^— xh  +  z^ 
2x^y-\-2xy^-{-2xyz 

xy^  +  jf — ^^^'  —  ^xyz  +  ^ 
xy^  +  !/^  +  y^z 


— xh — 2xyz — yh  +  z^ 
— xh —  xyz — xz^ 


—  xyz — yh  +  xz^  +  !^ 

—  xyz — yh — yz^ 


xz^  +  xz^  +  z^ 
xz^ -\- yz^  ■\- z^ 

0 


43  KEY    TO     DAVIES'    BOURDON.  [55-56. 

32.   a^— .yg + ^yh — 3yz-  ^:?    x—y-\-z  


yfi—xtyj^xh  I  xi-\-xtj—xz-\-y^—'-lyz-\-z^.  Am 

xiy—xh—y^  +  oi/z — ?>ijzi  +  2^ 

x?y—xy^-\-xyz  

— xh — xyz  +  .ry'^ —y'^  +  3^-2 — 'dyz^  +  z^ 

— xh-{-xyz—xz^ 


xy^—y^-{-Sy'h—2xyz  +  xz^—dyz^  +  z^ 

xy^—y^+  y~z 

— 2:17/2  -h  2yh  +  3-2;2 — dyz^  +  2^ 
—  2xyz  +  2yh         —2yz^ 

xz^—  yz^  +  ^ 
xz^ —  yz^  +  z^ 

FACTORING. 
Art.  46,  'pp.  55-50. 


1.  5a2  X  5a2  —  ha^  x  6aJ  +  5«2  x  W 

=  ho?  (5rt2  —  6a5  +  3&2).     ^».f. 

2.  3a2  X  J  +  3a2  x  3c  +  3a2  x  ^xy 

—  3a~(b  +  3r  +  Ga;?/).     ^z*^'. 

3.  2ac  X  4a2;  —  2cc  x  dx^  +  2ac  xc^y  —  2ac  x  lbct>(^x 

=  2ac  {4:ax  —  9x^  +  c^y  —  15«^A).     ANf<. 

4.  Qabc  X  4:abx  —  6abc  x  ha'b^c^y  +  6rt&r  x  (5««*'fZ  +  Gabr  x  1 

^  Qabc {4.(ibx  —  5rt^i*<'5?/  +  GaW'd  +  1 ).     .1  y/.--. 

5.  dab-^  X  a2  —  3aJ-2  x  2fl J-i  +  dab-^  x  4&-2c 

.     =  3ab-^  (a2  -  2^^!  +  U'^c).     A  us. 

6.  ix-^y-^  X  2?/-2  _  4,r-2//-i  X  Sx-^y-"^  —  4a;-22/-i  x  i.';-- 

=  4rt-27/-i  (2y-«  —  3.T-it/-i  —  4.i-2).     J  y/.«^. 


56.  J  FACTORING.  43 

Art.  47,  p.  56. 

7.  (3a;)2  +  2  x  3a;  x  4^/  +  (4y)2 

=  (3a;  +  4:y)  (3a;  +  4?/).     Jws. 

8.  {^xyf  —  2  X  2.f?/  X  xhf  +  (.t-y )2 

=  (2a;?/  —  x^y^)  {2xy  —  .r^^^).     Afis. 

9.  (3a;?/2)2  -  (2a?^?/)2  =z  {Sxy^  —  2j^y)  {6xy^  +  2xhj).     Ans. 

10.     (aa;)2  —  {hx'^f  =  {ax  +  bx-^)  {nx  —  bx"^).     Ans. 

n.     (ay-^'Y  +  2  («.?/-i>)  (bx-^)  +  (ia^m)2 

=  (fl!i/-°  +  bxr-^)  {ay-''  +  ^.<-'°).     Ans. 

12.  2ar-2  [(3a;-i)2  +  2  (3a;-i)  y  +  (^)2] 

=  2a; -2  (3a;-i  +  ij)  (3ari  +  v^.     J«.s. 

^rf.  48,  1),  56. 

13.  (a;)2  +  (7  +  10)  x  +  (7)  x  (10) 

=  (a;  +  7)  (a;  +  10).     Ans. 

14.  (a;)2  + (_7-10)a;+  (-7)  x  (-10) 

=  (a;  —  7)  (a-  —  10).     Ans. 

15.  (a;)2  +  (-7  +  10)a;+  (-7)  X  (10) 

=  (a;  —  7)  (a;  +  10).     Ans. 

16.  (a;)2  +  (7  -  10) a;  +  (7)  X  (-  10) 

=  {x  +  7)  (a;  —  10).     Ans. 

1  ;■.     {xf  —  (2J2)3  —  (a;2  +  2J2.C  +  4J2)  (a;  —  2^*2).     Ans. 

IS.     (.i;2)3  _  (3a)3  =  (.r*  +  3aa;2  +  9a^)  (.i-2  _  ;3a).     Ans. 

10.      (a;)2  +  (3  +  5)x  +  (3)  x  (5)  =  (a;  +  3)  (a;  +  o).     ^w*-. 

20.     x[(a;)2  +  (3-5)a;  +  (3)x(-5)] 

=  X  {x  +  3)  (.r  —  o).     ^1«5. 


44  KEY    TO    DAYIES'    BOURDON.  [57-58. 

Art.  49,  2>.  57. 

21.  (3a)3  —  (2J)3  =  (3a  —  U)  {%a?  +  Qah  +  452).    ^?^s. 

22.  (3a:)3  +  (%)3  =  (3:r  +  %)  (9a-2  —  12.r?/  +  16?/2).     ,4»s. 

23.  (aJ)*  -  (cr/-i)* 

=  («Z*  —  cd-^)  {ah  +  c^-i)  (a262  +  cH-^)-    Ans. 

24.  (a;-i)3  —  {yf  =  (.^-1  —  y)  (2-2  +  .1-1^  +  f~).     Ans. 

25.  (2aa;2)4  _  (^2^.)4 

=  (2aa;2  —  ^2^)  {^lax^  +  ^2^;)  (4a%2  +  yix'^),     Ans. 

26.  3a;2  (856.i^  -  y^)  =  ^x'  [{2b^x)^  -  (yY] 

=  3a:2  {2b''x  —  y)  {Wx^  +  Wxy  +  y\    Ans. 

Art.  51,  2>'  57. 

30.  2a  {5x  4-  4ryy  +  3b  {5x  +  4?/) 

=  (2a  +  35)  (5a:  +  4y).    ^w«. 

31.  a^  —  dx^  +  Sx  —  2  =  (x^  —  a;2  +  a;)  —  2.r2  +  2a;  —  2 

=  a;(a;2— a;+  1)  —  2  (a;2  —  a:  +  1) 
=  (a;  — 2)(a;2  — a;  +  1).     Ans. 

32.  a;2  {x  —  a)  —  a^  [x  —  a)  =  {x^  —  a2)  {x  —  a).     Ans. 

GREATEST    COMMON    DIVISOR. 

Art.  54,  i)p.  58-59. 

2.     Factoring,  a;2  —  a:  —  2  =  a'2  +  (1  —  2)  a:  —  2  x  1 

=  (a:  -  2)  (a:  +  1)  ; 
and  a^  +  2a:  -h  1  =  (a;  -F  1)  {x  +  1). 

.*.    g.  c.  d.  =  X  -\-  1.    Afis. 


8-62] 


GBEATEST    COMMON    DIVISOR. 


45 


3.  Factoring,        a:^  —  a^  —  (x  —  a){x  +  a) ; 

and  x^  +  a^  =  {x  +  a)  (x^  —  ax  -\-  a^). 

.'.    g.  c.  d.  =  a;  +  a.    Ans. 

4.  Factoring,  x^  +  o.i^  =  .r^  (a:  +  5)  ; 

and     .t2  ^  (5  —  1)  .T  —  5  X  1  =  (.»  —  l){x+  5) ; 
/.    g.  c.  d.  =  a*  +  5.     Anfi. 

5.  Factoring, 

a;2  +  (_  3  -  1)  a;  +  (-  3)  X  (-  1)  =  (x  -  3)  (x  -  1) ; 
and  a;2  +  (-  3  +  2)  a;  +  (-  3)  X  (2)  =  (.t  -  3)  (,i-  +  2) ; 

.'.     g.  c.  d.  =  a;  —  3.     ^??s. 

6.  Factoring,        {xf  +  2x  +  (1)2  =  (a-  +  l)  (a:  +  1)  ; 
and        x^  +  2x^+2x  -\-l  =  {x^  +  a;2)  +  (^2  ^  a;)  +  (a;  +  1 ) 

z=  a;2  (a-  +  1)  +  a;  (a;  +  1)  +  1  x  {x  +  1) 
=  (a-2  +  a:  +  l)(a;  +  l); 
.*.     g.  c.  d.  =  .r  +  1.    A)is. 


Art.  54,  p.  62. 


3. 


FIRST  OPERATION. 


a;3  +  3a,-2  +  4a;  +  12 

a;2  +  4a:2  -f-  4a;  -f    3 

—    a^  +9 


a;3  ^  4^2  ^  4a;  ^  3 


SECOND  OPERATION. 


—  o"2 


a;2  +  9 


—  X 


x^  +  4a;2  -|-    40*  +    3 
a;3  —    Oa; 

4x2  4.  13a;  +    3 
+  4a;2  _  36 

133-  +  39 
Rejecting  the  factor  in  tins  remainder,  we  have  a; +3. 


Of  -r!--ir 


46 


KEY    TO    DAVIES'    BOURDON". 


[62. 


THIRD  OPERATION. 

-a;2 

+  9 

X  +  3 

-a;2 

—  3x 

—  x  +  d 

3x+  d 

3a;  +  9 

.:     g.  c.  d. 

=  X 

+  3. 

A  lis 

FIRST  OPERATION. 


a^  +    a:2  +    x  —  3 

a:3  +  3a;2  +  5a;  +  3 


a;3  4-  3a;2  4-  5a:  +  3 


—  2x^  —  4:X  —  6 
Rejecting  the  factor  —  2,  we  have  a?  +  2x  +  S. 


SECOND   OPERATION. 


a^  +  Sxi+5x  +  3 
a^  +  2x^  +  3x 


x^  +  2x  +  3 


X   +  1 


x?  +  %x  +  3 

x^-\-^x  +  3    .-.  g.c.  d.  =  a;2_j_2a;-|-3.  ^/?5. 


FIRST  OPERATION. 


5. 


ofi  +  mx^  +  mx  +  1 


cfi  +\ 


a? 

4-1 

1 

mx^  -\-  mx 

ejecting  the  factor  mx,  we  have  x  + 1. 

SECOND  OPERATION. 

.^3+  1 

x-\-\ 

1 

X?  ^-^ 

a;2  — 2;  + 

—  a:2  +  1 

—  o^  —  x 

a;  +  l 

x-\ 

-1       .-.     f 

5.  C.  d.  : 

X  -y  \.    Ans. 


62.] 


GREATEST    COMMON     DIVISOR. 


47 


6. 


FIRST  OPERATION. 

6x^  —  7ax^  —  2Qa^x    j  3:?.'^  +  ax  —  4a^ 
ea?  +  2ax?  —    Sa^a:    j  3.r 


SECOND  OPERATION. 


3a;2  +    rto:  —  4^2 

3a;2  +  ^ax 

—  Zax  —  4^2 
_  3«a:  —  4<?2 


'ix  +  4« 


a: 


a 


g.  c.  d.  =  3x+4a.  ^««. 


7. 


FIRST  OPERATION. 

a:4  _  3a;3  ^  2a;2  +    a;  —  1  ,  a:^  —  x^ 
xi—    x^  —  2x^  +  2x 


2a;  +  2 


a:  — 2 


—  2a^  +  4:^2  —    x  —  1 

—  2x^+2x^  +  Ax  —  4: 

2xi  —  5a:  +  3 
We  now  multiply  the  second  polynomial  by  2. 

SECOND  OPERATION. 


2a.-3  —  2a:2  _  4^^  +  4 
2x3  —  5a:2  +  3a: 

3a:2  _  7.-y  4.  4 

6a:2— 14a;+  8 
6a:2  — 15a:+  9 


2.7-2  _  Qx  +  S 


X,  3 

(Multiply  by  2.) 


a;  —  1 

Note. — We  may  multiply  any  partial  remainder  by  a  monomial  factor. 
In  this  case,  we  multiply  the  first  partial  remainder  by  2. 


THIRD  OPERATION. 


2x2  _  5a;  +  3 

x-1 

2a? -2x 

2a:  — 3 

—  3x  +  3 

—  3a:  +  3 

•••    g.  c. 

.  d.  =  a:  —  1.     Ans. 


4B 


KEY    TO    DAVIES'    BOURDON. 


[62. 


FLRST   OPERATION 

a4  _  7a;3  _^    82;2  +  28a:  -  48 

a,-3  _  8a;2  +  19a;  - 

-14 

yA  _  8.r3  +  19.^2  _  14a; 

a;   +  1 

a;3  _  11.^^2  +  42a;  —  48 

a;3  _    8a;2  +  19a;  —  14 

—    3a;2  +  23a;  —  34 

Multiply  second  polynomial  by  3. 

SECOND  OPERATION. 

3a:3  _  24x2  +  57^;  _  42     1  -  3a:2  +  23a;  -  34 
Zt?  —  23a;2  +  34a;  | 


X 


—      a;2  +  23a:  —  42 

THIRD  OPERATION. 

—  3a;2  +  23a;  —    34 

—  3a?  +  69a;  —  126 


—  a;2  +  23a;  -  42 


46a;  +    92  =  —  46  (a;  —  2) 

FOURTH  OPERATION. 


—  ar!  +  23a;  —  42 

—  a;2  +    2a;    

21a:  —  42 
21a;  —  42 


X—    2 


-a; +  21 


9.      Multiply  the  first  polynomial  by  3. 

FIRST   OPERATION. 


12:<;4  ^  27a;3  _]_    6a;2  —    6a; 
12ar^  +  20a;3  _    4-^2  +    8a; 


12 


g.  c.  d.  =  a;  —  2.  Ans. 


3a;3  +  5a:2  —  a;  +  2 
4.a;,  7 


7a;3  +  10a;2  —  14a;  —  12     (Multiply  by  3.) 

21a;3  +  30.^^  —  42a;  —  36 
21a;^  +  35a:2  _    7a;  4-  14 

_    5a;2-35a;  — 50  =  -  5  (a?  +  7a;  +  10) 


62.J 


GREATEST    COMMON    DIVISOR. 


49 


SECOND  OPERATION. 


Sa^  +  21a:^  +  30a; 

—  16a;3  —  31a;  +   2 

—  16a;2  —  112a;  —  IGO 


a;2  +  7a;  +  10 


3a;  —  16 


81a;  +  162  =  81  (a;  +  2) 


THIRD  OPERATION. 


x^  +  7a;  +  10 
x^  +  2x 

5x  +  10 
5x  +  10 

0 


a;  +  2 


a;  +  5 


g.  c.  d.  =  X  +  2.     Ans. 


(10.) 

FIRST  OPERATION. 


x^—2x*—6a^-\-   4a;2+J3a;  +  6 


x^—2x*—6x^-\-4:Z^-{-13x  +  (i 


1 


5a;4  _|_  6a;3— 12.a;2— 22a;— 9 
Multiply  the  second  polynomial  by  5. 


SECOND  OPERATION. 


5a'5— 10a;*— 30a;3+   20a;2+   65a;  +  30 
5a-5+   6a^— 12a;3—   22.r2—     9x 


5a;*  + 6a;3— 12a-''— 22a;— 9 


.7-.     -  8 


—  16a;^— 18a;«+   A2x^+   74a;  +  30    (Multiply  by  i) 
_  40.t4 — 45a;3  ^  1  oSa:^  + 1 85a-  +  75 
_40a:4_4a,^_^.   96a;2  +  176a;  +  72 

3ar5+     9^2^     9a;+   3  =  3(2-3  +  3a^  +  3a-+l) 


50 


KEY    TO    DAVIES"     BOURDON. 


[62. 


THIRD   OPERATION. 


a;3  _1_  3a;2  _j_  32;  ^   1 


5a;  — 9 


5a4  ^    6:^3  —  12a:2  —  22a:  —  9 
5a^  +  15^  +  15a:^  +    5x 

—  9x^  —  27a;2  —  27x  —  9 

—  9x^  —  27a;3  —  27a:  —  9 

.-.     g.  c.  d.  =  a:^  _j_  3a;2  _|_  3^;  _|_  1, 
a:3  +  3a;2  +  3a;  +  1  =  {x  +  1)  {x  +  1)  (a;  +  1).    Ans. 

11.      Multiply  the  first  polynomial  by  3. 

FIRST   OPERATION. 


12a;5_  sx*—22'x^—     6a:'~—  6a:—  2 
12a;5^_   6a:4— 54a:3-|-     9a^— 15a: 


4a:<  +  2.r»— 18a;2  +  Sa;— 5 


3a-,      -  7 


— 14a;4^32.^■3_  15a-2+    9a'—   2     (Multiply  by  3.) 
— 28a;4  +  64a;3_  302-2+ I8.r—  4 
— 28a;^— 14a;3  +  126a:2— 21a:  +  35 

78a:3_i56a;2  +  39a;_39  =  39  (2a:3_42;2  +  a;_i) 


SECOND   OPERATION. 


4a:«  +    2a:3  _  iga;^  +  3a;  —  5 
^x*  —    8a;«  +    2a^  —  2x 

10a;3  _  20a:2  +  5^;  _  5 
102;3  _  30^2  +  5a;  —  5 


2x^  _  4a;2  +  a;  —  1 


2.r   +  5 


.-.     g.  c.  d.  =  2x^  —  4a;2  -\-  x  —  1.     Ans. 
12.      Multiply  the  first  polynomial  by  3. 


FIRST   OPERATION. 

60a-«-36a;5+  48.t«_  453^+     43x--  45a;  +  13 
ri0a!«-36x5  +  188.T^-  84a;34.  ii22;« 


15ar'-9a:8  +  47a;'-31a;  +  28 
'  4a;^     -28 


-140ar»+  39a;»-     703?-^-  45a;  +   13        (Multiply  by  3.) 
-430a;<  +  117a'3_  210/^-135.r+  36 
-420a;*  +  2533;'-  ISlGx'^  +  5883^-784 
-  135a;'  + 1 1 0()a;''  -  733a;  +  830 
Multiply  the  second  polynomial  by  9. 


m] 


GREATEST    COMMON    DIVISOR. 


51 


SECOND  OPERATION. 


Vdox*-      81x3+       423x^-       189x+       253 

-135x3  +  1 106x2  -  733x  +  820 

135j-^-  1106x3+       723x--       830x 

-X,     -205 

1035^3-      300x2+       (J3ia;+       253 

(Multiply  by  27.) 

27675x3-     8100x2+   17037x+     6804 

27675x3-226730x2  +  148215x- 168100 

218630x--131178x  + 174904  =  43736  (5x2-3x  +  4). 

THFRD   OPERATION. 

5^  —  3a:  +  4 


—  135a^  +  1106a;2  —  723a;  +  820 

—  135a;3  ^      gi^i  _  ipsa;  

1025a;2  —  615a;  +  820 
1025a;2  _  615a;  +  820 

0 
/.  g.  c.  d.  =  5a;2  _  3a;  +  4.  Ans. 


—  27a;  +  205 


FIRST    OPERATION. 

13.       ccS  _^_  3a;3  _  2a;2  +  14a;  +  21 
x^  -I-    2:3  _  22;2  +     Cjx 


2-4  -^  x^  —  2x  +  ^ 
X    —   1 


_  a4  _  42;3 

+    8a;  H-  21 

—  a;*             — 

a;2  +    2a-  —    6 

—  4a;3  _^    a;2  +    6a;  -f  27 
Multiply  the  second  polynomial  by  4. 


4a;* 
4a;4 


a:3 


SECOND  OPERATION. 

+    4a;2  —    8a;  +    24     |j- 4a;3  _^  a;2  ^  ga;  +  27 
—    6a;2  —  27a; 


X,      -1 


x^  +  10^;^  +  19a;  +     24     (Multiply  by  4) 
4a;3  ^  40x2  +  76a;  +    96 


4a;3_ 


X' 


6x—    27 


41a;2  +  82a;  +  123  =  41  (a;2  +  2a;  +  3) 


52 


KEY    TO    DAVIES'    BOURDOIf. 


[63. 


THIRD  OPERATION. 

4a;3  _  8^.2  _  122: 


4:X  +  9 


9^^  +  iSx  +  27 
9x^  +  18.^  +  27 

0 

.-.    g.  c.  d.  =  x^  +  2x  i-  3.     Ans. 

)        14.      We  first  find  the  greatest  common  divisor  of  the  first  and  second 
polynomials  : 

FIRST  OPERATION. 


x^—    9a:2  +  26a:  —  24 
a;3  _  ioa.2  ^  31a;  —  30 

x^—    5x  +    6 


0:3  _  ioa;2  +  31a;  —  30 


SECOND  OPERATION. 

a^  —  5a;  +  6 


X 


x?  —  10a;2  +  31a;  —  30 
X?  —    5x^  +    6x 

—  5a;2  +  25a;  —  30 

—  Sa:^  +  25a;  —  30 

0  .-.     g.  c.  (i.  =  X?  —  bx  +  6. 

We  next  find  the  greatest  divisor  of  this  result  and  the  third  polyTiomial. 


THIRD   OPERATION. 


ofi  —  lla;2  +  38.r  —  40 
ofi—    5a;2  +    6a; 


2-2  _  5a;  _,_  6 


X   —& 


—  6a;2  +  32.a;  —  40 

—  6a;2  +  30a;  —  36 


2a; 


4  =  2  (a;  —  2) 


63.] 


GREATEST    COMMON    DIVISOR, 


»3 


FOUBTH  OPERATION. 

x  —  2 


x  —  3 


x^  —  5x  -}-  a 
a?  —  %x 

—  3a;  +  6 

—  3a;  +  6 

0        /.    g.  a  d.  =  a-  —  2.    Ans. 

15.      We  find  the  greatest  common  divisor  of  the  first  and  second 
polynomials. 


FIRST   OPERATION. 


a*  +  lOa^  +  30a;2  —  \0x  —  31 
a^  —  10a;2  +    9 


x^  —  10.^2  +  9 


10a:3  ^  30a;2  _  lo^^  _  30  =  10  {x^  +  dx^—x—%) 


xi  _  10a;2  +  9 

—  3.t3  —    9a;2  +  3a;  +  9 

—  3a:3  —    9a;'-  +  3;r  +  9 


SECOND  OPERATION. 

x^  +  3a2  —  a;  —  3 


X  —d 


0 


g-  c.  d.  =  a;^  +  3a;2  —  a;  —  3. 


We  then  find  the  greatest  common  divisor  of  this  result  and  the  third 
polynomial 


THIRD   OPERATION. 


a4  +  4a:3  _  22a^  —  4a;  4-  21 
a:4  4.  3a;3  _      x^  —  3x 


a^  _  2la;2  — 

X  +  21 

a^  +    3a;2  — 

X—    3 

a;3  +  3a;2  —  a;  —  3 


a;  4-  1 


—  24.^2 


4-  24  =  _  24  (a-2  —  1) 


M 


KEY    TO    DAVIES'    BOURDON". 


[63. 


FOURTH  OPERATION. 


2^  +  3a;2  - 

X  - 

-3 

a? 

— 

X 

3a;2 

-3 

32;2 

-3 

a?  —  1 

a;  4-  3 


0  .-.    g.  c.  d.  =  x^  —  1.  Ann. 

16.  This  is  best  solved  by  factoring,  which  can  be  done  by  formulas 
(7)  and  (8),  p.  55,  and  formula  (3),  p.  41. 

x^  —  a^  =  {x  —  a)  {x  +  a)  (x^  +  a^) 
7^  -{-  a^  z=z  {x  -\-  a)  (a-2  —  ax  +  a^) 

X^  —  «2  —    ^  J,.  ^   (f"^  ^y.  _  f(^^ 

We  see  that  x  +  fi  is  the  only  factor  common;  hence  it  is  the  required 
common  divisor. 

17.  We  find,  as  in  the  last  example,  that  «  +  &  is  the  greatest  common 
divisor  of  the  second  and  third  polynomials.  But  a  +  b  will  divide  a'"  +  b'. 
according  to  principle  second,  p.  54;  hence,  a +  6  is  the  greatest  common 
divisor  required. 

18.  Find  the  greatest  common  divisor  of  the  first  and  third  polyno- 
mials. 

FIRST  OPERATION. 

%x^  —    "Ix^y  +  bxy^  —  y^ 
Zx^  +    bx'^y  +    xy^  —  y^ 

—  l%x^y  -f  ^xy"^ 


3a;3  +  bx^y  +  xy'^  —  y^ 


=  —  4^?/  (3^^  -  y) 

SECOND   OPERATION. 

Bx^  +  5x^y  +    xy^  —  y^    3x  —  y 

3x^  —    x^y  x^  -f  2xy  -f  y^ 


6x^y  +    xy^ 
%xhj  —  2xy' 


Sxy^  —  y^ 

dxy^  —  y^ 

0 


.  c.  d.  =  3a;  —  y. 


63-64.]  LEAST    COMMON    MULTIPLE,  55 

We  now  arrange  this  result  and  the  second  polynomial  with  respect  to 
y,  and  proceed  as  shown  in  the 


THIRD  OPERATION. 


—  2/3  +  ^xy^  +  x^y  —  3r»3 

—  y3  -|.  Sxy^ 


—  y  +  Sx 


yi  —  a;2 


+  3^y  —  3a^ 
+  Qfiy  —  Sa^ 

0 
.-.    g.  c.  d.  =  dx  —  y.    Ans. 

LEAST    COMMON    MULTIPLE. 
Art.  58,  p.  64. 

3.  Factoring,  we  have, 

2  x3a;,        2(a;  +  l),         2  (a: +  2); 
whence,  by  the  rule, 
1.  c.  m.  ==  2  X  Sx{x  +  1)  (a;  +  2)  =  6a;  (.r^  +  3a:  +  2).  A7is. 

4.  Factoring,        a,        a  +  2b,        {a  —  2b)  (a  +  2^*) ; 

whence,        1.  c.  m.  =  a  (a  +  2b)  {a  —  2b) 

=  a{a^  —  4J2).     Anf<. 

5.  Factoring,         {x  -  I),        2(a;  +  l),        2  {x^ -^  1) ; 

whence,         1.  c.  m.  =  2  (a;  —  1)  {x  +  1)  {x^  +  1) 
=  2  (a;*  —  1).     Ans. 

6.  Factoring,     2xa;(a;  +  l),     3xa;(a:— 1),     x{x—l){x  +  l); 

whence,        1.  c.  ui.  =  2  X  3a:  {x  +  1)  {x  —  1) 
=  6x  (a;2  —  1).     Atis. 


56  KEY    TO    DAVIES'    BOURDON".  [65. 

Art.  59,  p.  65, 

2.  The  two  polynomials  may  be  written, 

3a;2  —  2x  —  (3a;  —  2)  =  x  {3x  —2)  —  l{dx  —  2) 

=  (x  -  1)  {3x  —  2) ; 

4a:3  _  4a;2  —(x  —  l)  =  4:X^  (a;  _  i)  _  i  (^  _  1) 

=  (a;_l)(4a;2-l); 

,                  3x^—  5a;  +  2 
.-.    1.  c.  m.  = ~ —  X  (4a;3  _  4^2  _  ^^  ^  1) 

*C  "^  J. 

=  (3a;  —  2)  (4a;3  —  4.r2  _  a;  +  1).     Ans. 

3.  The  greatest  common  divisor  of  the  two  polynomials  may  be  found 
as  in  the  last  example;  it  is  2a;— 1.  The  quotient  of  the  first  polynomial 
by  this  result  is  3a; +  1.     Hence,  by  the  rule, 

1.  c.  m.  =  (3a;  4-  1)  (2a;2  +  3a;  —  2).     A71S. 

Art.  60,  p.  65. 

4.  The  greatest  common  divisor  of  the  first  two  polynomials  is 

a:2  —  5a;  +  6  =  (a;  —  2)  {x  —  3). 
Dividing  each  polynomial  by  this,  we  see  that  the  two  can  be  written, 
(a;  —  1)  (a;  —  2)  (a;  —  3)     and     {x  —  2)  (a;  —  3)  (a;  —  4) ; 

hence  (Art.  58),  their  least  common  multiple  is 

(a;— I)  (a;— 2)  (a;— 3)(a;— 4)  —  c^—lOa^  +  dhx^—h^x  +  ^A; 

hilt   this   result  is  divisible  by  the  third  polynomial.     Hence,  the  least 
couiniou  multiple  of  the  three  polynomials  is 

(a;_l)(:r-2)(a-  — 3)(a;  — 4).     Ans. 

NoTK. — The  following  examples  maybe  solved  in  the  same  way,  but 
tliey  arf  more  easily  solved  by  the  rule  in  Art.  58. 


65-69.]  TRANSFORMATION    OF    FRACTIO^TS.  '^7 

5.  The  greatest  commou  divisor  of  the  first  two  polynomials  is 

.^  +  3a;2_a;— 3  =  x^x  +  3)  —  1  {x  +  3)  =  (x^—l)  (.t-^-). 

Hence  the  first  polynomial  can  be  written  (a-^— 1;(.t  +  3)  (.r— 3),  and  the 
second  polynomial  can  be  written  (a;^— 1)  (^  +  3)(x  +  7).  If  we  divide  the 
third  polynomial  by  .i- — 1,  we  find  for  a  quotient 

a;2  +  ix  -  31  =  {X  4-  7)  {x  -  3) ; 

hence,  the  third  polynomial  may  be  written  {x^—l){x+7){x—3).  By  the 
rule  of  Art.  58,  we  have 

1.  c.  m.  =  {x^  —  1)  (a;  +  3)  (x  -  3)  {x  +  7) 
=  (a;2  _  1)  (a;2  _  9)  (a;  +  7).    Ans. 

6.  Factoring,  we  have, 

For  the  first,  {x  —  a)(x  —  b); 

For  the  second,  {x  —  b)  {x  —  c); 

For  the  third,  {x  —  c)  (x  —  a)  ; 

.:    1.  c.  m.  =  (x  —  a)  (x  —  b)  {x  —  c).    Ans. 

7.  Factoring,  we  have. 

For  the  first,  2xS{a^-ab  +  b-^){a-b){a-b)ia-b){a-b); 

For  the  second,        3  x  3  (a^  +  b^)  {a  +  b)  (a-b)  (a-b)  (a-b); 

For  the  third,  2x3x3  (a  +  b)  {a +  b){a  +  b){a-b){a-b){a-b) ; 

.•.     1.  c.  m   = 
2x2xdxS{a^-ab  +  ¥)(a^  +  b')(a-b){a-b)(a-b)(a-b){a  +  b)ia  +  b)(a  -  b) 
=  36[(a2  +  6^)(a  +  6)(a_5)]  x  la'-ab  +  ¥){a-b}]  x  [{a  +  b){a  +  b)ia-b){a-b)] 
=  36  (a-'-b*)  {a^-¥)  {a'^-b^y.    Ans. 

TRANSFORMATION    OF    FRACTIONS. 
Art.  71,  p.  69. 

^      Qa^b  +  ^ac  _  3a{^ab  +  c)   __  2ab  +  c 

^ab+3ad  ~  "3a(3J  +  fO    ~    U  +  d'       ^^' 
25bc  +  5bf  _  5b(5c+f)   _  oc  +  f 
'     rob^+Vob  -  5b{7b  +  3)  -  75  +  3'        ^^*- 


58  KEY    TO    DAVIES'    BOUEDON.  [69-70. 

36a^  +  12abf  _  Uai  {3a-\-f)  _  3a-}-/ 
•  84aJ2     "    ~      12abx7b      ~  '  U    ' 

12acd—4:c^  _  4:cd{Sa—d)  _  Sa—d 
^'     12cdf+  ^M  ~  I^J(3/+^)    "~  3/+c'        "*• 

18«V — 3acf  _  3«c(6ac— /)   _  '6ac~f 

7.     The  greatest  common  divisor  of  the  terms  is  %x  —  5. 
6a,-2  _  7a-  _  20 


2x  —  h 

4a^  —  27a;  -I-  5 

3a;  -f-  4 
"     2a;2  +  5a;  —  1 


=  3a;  -f-  4  ; 

=  2a;2  +  5a;  —  1 ; 

•     Ans. 


8.  The  greatest  common  divisor  of  the  terms  is  a;  —  1. 

a;2  +  a;  _  2  ^  2a;2  —  3a;  -^  1         ^         ^ 

—^ — - —  =  a;  +  2; -^-   =  2a;  —  1 ; 

a;  —  1  X  —  1 

a;  H-  2         . 
•••     2^31-     ^^^- 

9.  The  greatest  common  divisor  of  the  terms  is  .r  —  3. 

3a;3  —  22a;  —  15 


X  —  3 
5a;4  _  i7,^;3  ^  ig^; 

a;  —  3 

3a;2  _|_  9a;  _|.  5 


=  3a;2  +  9a;  -I-  5 ; 
=  5a;3  _  2ar!  —  6a; ; 
Ans. 


5a;3  —  2a;2  —  6a; 

10.     The  greatest  common  divisor  of  the  terms  is  x™  —  1 

a;3m  _|_  ^2m  _  2 


X^  —  \ 
a«m  4.  ^m  _  2 

a;2m  _^  2a-m  -f-  2 


a;°»  +  2 


=  a;2m  +  2a;m  -(-  2; 
=  a;°i  -I-  2 ; 


70-71.]  TBANSFORMATIOX    OF    FRACTIONS. 


59 


Art.  72,  p.  70. 

^      5  X  3a:  +  (2a;  -  7)         17a;  -  7        . 
2. 'ir-'^ =  — Ans. 


Sx 


dx 


a  —  (x  —  a  —  1)         2a  —  X  +  1        . 
a  a 

(1  +  2a;)  X  oa;  —  (a;  —  3)   _  lOa;^  +  4a;  +  3 
bx  53" 

{Zx  —  1)  (3a  —  2)  —  (a;  +  a)  _  9aa;— 4cf  — 7a;  +  2 


4. 
5. 


3a  — 2 


3«  — 2 


A71S. 

.   Ans. 


g      (aa:  +  4)  {x  +  y)  —  {ax^-y)  _  axy  +  ^x  +  by      ^^^^ 

x  +  y  x+y 

^      3(a;2  -  1)  +  3            3a;2  . 

7.     -^^ — ^ — '-^ —  =  ~ T-     Ans. 


8. 


x^  —  1  x^  —  1 

(a2  —  ax  +  a;3)  {a  +  x)  +  2a,-3  _a?  +  3a,-3 


a  +  X 


a  +  X 


Ans. 


1.     aa;  +  a~  I  a; 
a  -\ Ans. 

X 


Art.  73,  p.  71. 

2.     ah  —  2a^  |  b 
2a2 
o 


10a:2  —  5a;  +  3  |  5a; 

I         ~,        3        , 
2a;  —  1  +  ^--     Ans. 
6x 


4.     x^  -\-6x-^  12y  —  5?/2 
a;2  +  6a;  —  2a;_?/ 

2xy-{-l2y — 5y^ 
2xy-\-12y — 4y^ 


x—2y  +  6 


x  +  2y  — 


r 


x  —  2y  +  6 


Ans. 


.yi 


60 


KEY    TO    DAVIES'    BOURDOX. 


[72-73. 


5. 


6. 


a;2  +  ■<^ 

X^  —  X 

2x 
2x 


x—l 

..  +  2+- 

2 

— 

1 

Ans. 


3 

Aax  —  2x^  —  «2 
4aa;  —  2x^ 

-«2 


2a- 

-  X 

2x- 

a2 

2a 

— 

X 

•    Ans. 


1. 
2. 


4. 
5. 


Art.  75,  PI).  73-74. 

axe      (a  +  h)b  ac     ah-\-^        . 

-T — ,    ^ — T-^,    or    — ,    — A71S. 

be  be  be         be 

1  X  3  X  {a  +  x)      a^x2x  (a-i-x)      (a^  +  x'^)  x  2  x 3 
2x3x(a  +  4'     2  X  3  x  (rt  -frj '     2x3  x  {a  +  xY' 


or 


3rt  +  3:r      2a^+2d^x      Ga^  +  6x^ 
6a  +  6a: '      6a  +  6a;   '     6a  +  6x 

axaxxe       {e—b)x{a  —  b)xe    bx{a — b)xax 


{a — b)xaxxe^     (a  —  b)xaxxc    '(a — b)xaxxe' 
ac^ — abc — bc^  +  l^c     a^bx — ab^x 


or 


a^ex 


c^cx — obex 


cfiex — abcx^ 

1  X  (a  +  a;)  1 

{a-\-x)  {a—xY    cfi — a;2' 

1  X  {a—x)  1  X  a  (a  -f- a-) 


a-\-x 

or , 

w — x^ 


'    a^cx — abcx 
1 


Ans.. 


a^ — 7? 


Ans. 


c^{a—x){a-{-xy    axa^xia—x)  (a  +  x)' 
2xa 


axa^{a—x)  (a  +  x)' 
a—x  a{a  +  x) 


or 


2a 


a^{a^ — x"^)'     n^  (a^ — x^) '     a^  (a^ — x^) 


AnsI 


Ans. 


74-75.]  ADDITION    OF     FRACTIONS.  61 

6.  The  least  commou  multiple  of  the  denominator  is  (a  +  1)  (a— l)''^. 
We  divide  this  by  each  denominator  and  then  multiply  both  terms  of  the 
corresponding  fraction  by  the  quotient. 

xx{a-\-l)(a  —  l)  yx{a—l)    

(«_1)  X  {a  +  'l)  («-!)'     («  +  i)  [a-l)  X  (a-1)' 

z  X  («  +  l) 
i 1 or 

(a-l)  (a-l)x(«  +  l)' 

x(a^-l)  y{(i  —  ^) ^(^  +  1) 

(,,  +  l)(«_l)2'     («  +  i)(«2_l)'     ^a  +  l){a-lf 

7.  The  1.  c.  m.  is  ;r'-l  =  (.r-1)  (a-'  +  if  +  1;.     Hence,  as  before, 

{x+l){x^-\-x  +  l)  {X^  +  1){1--1)  ^  ^y 

(a;-l)xia;M-^-  +  l)'     (a:-l)  x  (a;^ -f. -*■  +  !)'    ^-^' 

(x^\){x^-\-x^V)       Or^  +  l)(^-l)  ^  .^,. 

a;^-!  '  a.-3-l        '     a;3_i 


ADDITION    OF    FRACTIONS. 

Art,  76,  pp.  7^-75. 

2.      Add  the  entire  parts  by  themselves  and  then  add  the  fractional 
parts  by  themselves  and  take  the  sima  of  the  results. 

2ahx  —  dcx^        . 

=  a  +  c  ^ 7 Ans. 

be 

^      X      X      X        6x      4:X      Sx        13x  ,    x 

3      --I 1--  = 1 1 =  -.-  =  ^  +  TK*     Ans, 

"^^     2^3^4        12^12^12         12  ^12 

7(a;  _  2)  _f_  3  X  Ax        19a;  —  14        . 
4.     -i ^-! = A71S. 

^       .     .  Ax^xx  ,   aix+a)        .      ,   Aofi^-ax  +  a^       . 

6.     4a;  +  — h    \       '  =  Ax  -i ^ —  •     Ans. 

2ax  2ax  2ax 


62  KEY    TO    DAVIES'    BOURDOIir.  [75-76. 

''     {a  +  h)  {a-b)  ^  {a-\-b)  {a-b)  ~  ifi-¥ 

1  X  {a  —  .r)        1  X  {a  +  x)  1  x  2a      _  4a 

9.      The  second  fraction  is  the  same  as  ^  ~  ^^'^_     Hence, 

a  —  X 

x{a  +  x)       x^—bax         2a^    _  2x^—4:ax-\-2a'^ 
a — X  a — X         a — x  a — x 

=    ^  ~'  '-^^^-^  z=  2(a—x).     Ans. 
a—x 

(a—x)  {a  +  b){a  +  x)  c{a—b)  {a-\-x) 


10, 

{a—0)  {a  +  O) 

+ 


{a—b)  {a-^b){a-\-x)       (a—b)  {a  +  b)  (a -j- x) 

d{a—b){a  +  b) 


{n-b){a  +  b){a  +  x) 

(g^ — ax^ + (M — bx^)  +  (a^c —abc-{-  acx — bcx)  +  {da^ — dtp') 
a^  +  o?x — ab — JPx 

_  a^  +  a^{b-\-c  +  d)  —  a{x^—cx  +  bc)—b{x^^cx+M) 
~  a^  +  d^x — ab^ — b^x 


SUBTRACTION    OF    FRACTIONS. 
Art.  77,  p.  76. 

„      27a;       14a;        13a; 

dx  +  ad       be  _  dx  +  ad  —  be 
bd  bd  ~  bd 

24a;  +  8a  _  lO^a;  +  35b  _  2ix  +  8a—10bx—36b 

40b  406        ~  m  ^^^' 


76.]  -    ■  SUBTRACTION    OF    FRACTIONS.  63 

5.      This  may  be  written 

_  X      X  —  a       ,.         cx      hx  —  ab 

,     3^--^  +  4  +  ^=^^  +  S  +  -S^ 

rr  +  hx  —  ah        . 

=  2x  -\ -. Ans. 

be 

1.      This  may  be  written 

113 


x~l       x  +  2       {x  +  2)2 
{x  +  2)2  {x  -  1)  {x  +  2)  3  (a:  -  1) 


(x  —  l){x-\-  2)2       (.r  -l){x  +  2)2       (x  —  1)  {x  +  2)2 
_  {x  +  2)2  —  (x  —1)  {x  +  2)  —  3  (a:  —  1) 

-  (:^;  _  1)  (a;  4-  2)2 

9 


(a;— 1)  (a;  — 2)2" 
2.      This  may  be  written 

5  1  24 


Ans. 


2  (a:  +  1)       10  (a;  —  1)       5  (2a;  —  3) 
5  X  5{x-  1)  (2a;  +  3)  1  x  (a-  +  1)  (2a:  +  3) 


"  2(a;  +  1)  x5(a;-l)(2a;-l-  3)       10  (a;— 1)  x  (.r+1)  (2a-  +  3) 

24  X  2  (a:  +  1)  (a-  —  1) 
5  X  (2a;  +  3)  (a-  —  1)  (x  +  1) 
_  ^5  (2a;2  +  a;  —  3)  -  (2a;2  +  5a:  +  3)  -  48  (a;2  -  1) 
~  10(a:2- l)(2a;  +  3) 

*      —  20a;  —  30 _  2a;  —  3   

~  10  (a;2  -  1)  (2a;  +  3)  ~  (a;2  -  1)  (2a;  +¥)' 
3.      This  may  be  written 

3  +  2a;      a;2  —  16a;  _  2  —  3a; 
2  —  a;"^    4  —  a;2  2+a; 

_  (3  +  2a;)  (2  +  a?)  x^  —  Wx (2-  3a:)  (2  -  x) 

'  '(2  -  a;)  (2  +  x)   "^  (2  -  a;)  (2  -f-  x)        (2  -  x)  (2  +  x) 
_  (6  +  7a;  +  2a;2)  +  (a;2  —  16a;)  —  (4  —  8a:  +  Sar^) 

(2  —  a;)  (2  +  x) 

2  —  X  1 


(2-x){2  +  x)        X +'2 


Ans. 


64  KEY    TO    DAVIES'    BOURDON".  [76-78. 

4.  This  may  be  written 

3  7  20:c  — 4 

1  —  2x  ~  1  +  2x  ~  1  —  4a;2 
_  3  (1  +^       7  (1  —  2a:)  _  20a;  —  4 
~"     1—1^  1  -  4.^2  1  -  4a;2 

_  3  +  6a;  -  (7  -  Ux)  -  (20a:  -  4)  ^  _0__  ^  ^      ^^^^ 
—  1  —  4a;2  1  —  4a;2 

5.  Reducin<T  the  fractions  to  their  common  denominator  a*—b*, 

(ft  -  h)  («2  +  ^•^)       ^>  (»2  +  52)  _  a  ja'  -  h^) 
a'^  —  b^  ^     «4  —  ¥         '  a"  —  ¥ 

_  (gs  4-  ah-i  _  a^h  -  Z>3)  +  (a~b  +  ^^3)  -  {af"  -  ff&^) 


2a&2 


a* -6* 


^»c*. 


MULTIPLICATION    OF    FRACTIONS. 

Art,  78,  p.  78. 

2x       dab       Sac        ISa^cx  . 

a         c         2o  Zaoc 

g  ab±^bx^a^ah_±^x^     ^^^^ 
a  X  X 

(^__52)^+J2)  ^  ^_|4  _   ^^^^ 

ax  +  a;  +  1   a;  -  1  _  (ga;  +  ^  +  1)  (a^  -  1)   ^,^, 
^'      «     ^  «  +  J  -      ff  {a  +  J) 

Note.— If  there  are  any  common  factors  in  the  numerator  and  denom- 
inator of  the  indicated  product,  strike  them  out. 

d2—ax-\-ax        g2_a;2    _        ^2  {gj—x^) 

^-  ^-a;       ^  a;(a;  +  l)  ""  a;(«-a;)x(a;4-l) 

a;  (a:  +  1) 


78.]  MULTIPLICATIOX    OF    FE  ACTIONS.  65 

{a  +  b)  X  X  {a  —  b)        x  (a  +  0) 

{a  -b){a^-b)x{a  +  b)  _    {a  +  bf 
'•  4  (a  -  6)  (a  -  b)  -  4.(a-b)'     ^'''' 

9       a;  (.r  +  y)  x  {^  -  f)    ^      a;"  -  3/'    .     ^,^^ 
try  (X-  +  i/)  X  (a;2  +  if)        y  {x^  +  f) 

daxx{a—x){a  +  x)y,b{c  +  x)y.{c—x)  _  dx 
4:byx{c—x){c  +  x)xa{a+x)x{a—x)~4:y 

M    x)  -\-  X  1 

11.      The  last  factor  may  be  written  — ^  or .     lli-nce, 

1  —  X  1  —  .« 

(1  -x){l+x)  X  {I-  y)  (1  +  y)  X  1  ^  l-.y   ^^^^ 
(1  +  y)  X  a;  (1  +  :»)  X  (1  —  a;)       a; 

x  {a  —  x)  X  a  {a  -^  x)  ax 

(a  +  x)  {a  -\-  x)  X  {a  —  x)  (a  —  x)  ~  {a  +  x)  {a  —  x) 

ax 


12. 


13. 


a^  —  x^ 


{a  -b){a  +  b)  (a^  +  b^)  x  (a  -  b)  _   ^  ^ 
{a  —  b)  {a  —  b)  X  a  {a  +  b)         ~ 


a 

=r  (I  -\-  -  ■     Ans. 
a 

14.  The  first  factor  may  be  reduced  to  the  form  "^ ^ ,  and  the 

second  factor  may  be  reduced  to  the  form   -^ •    Hence, 

x^  —  ax  +  a^      x^  -\-  ax  +  a? ir'  +  a^x^  +  a^ 

=  ^  +  «^  +  l-     ^^^- 

15.  Proceeding  as  before,  we  have, 

01^  —  x  +  \        x^  +  X  +  \         xf^  -\-  x^  +\  .,       ^         1 

1  x^  x*  x^ 


66  KEY    TO    DA  VIES'    BOURDON.  [80. 


DIVISION    OF    FRACTIONS.     : 
Art.  79,  i}p.  80-Sl. 

x-\-\       2x  _  x  +  1       3   _  3  (a;  +  1)  _  a;  +  l 
^-     "e^^T-^e"  ^2x-       13a;       -    4a;  •  ^'^• 

X  XX  2  2  . 

c;  —  l       2       x  —  1      X       x  —  \ 

X  —  h      'Sex       x—h       M        X  —  b       . 
8cc?         4c?  8c(l        dcx        bc^x 

Note. — In  cases  like  the  following,  we  indicate  the  operation,  factor- 
ing when  possible,  and  then  strike  out  all  factors  common  to  the  terms  of 
the  indicated  fraction. 


6. 


8. 


(a;  -b){x-{-  b)  [x^  +  ^)  X  (a;  -  Z>)  _  ofi  +  l^ 
{x  —  b){x  —  b)  y.  x{x  -[-b)        ~       x 

=  X  A Ans. 

X 

{ax  -  1)  X  (1  -  a;)  (]  +  x)  ^  (ax  -  1)  {x  +  1) 
(1  —  a;)  X  a  ~  a 

aa;(l  +  a;)  —  a;  — 1      . 
—  — 1 — 1 — i Ans. 

a 

(g  +  1)  X  (1  -  a)  (1  +a)  _  (a  +  1)  x  -(ff-l)(l  +  a) 
(fl-1)  X  (1  +  «)  (a-l)(l  +  a) 

=  —  (a  +  1).     .4ns. 


x{a  —  x)  X  (a  —  x){a  +  x)         {a  —  x^ 
{a  -{-  x)  {a  +  x)  X  x^  x  {a  +  x) 


80i]  DIVISION    OF    FRACTIONS.  6,7 

Aa(a-b)x  {a-b){a  +  b)  _   2  {a  -  bf^ 
b{a  +  b)  {a  +  b)  X  ^ab       ~  3J2  (a  +  b) 

11.     ^_fLx  (£+_^-)___  ^  ____% ^.     Ans. 

[y  +  ^)  if  —  y^  +  ^^)  X  y     y^  —  xy  +  ^^ 

12.  In  this  and  the  following  examples,  let  the  dividend  and  divisor 
be  reduced  to  the  form  of  simple  fractions  and  then  proceed  as  before. 


7^j\-_f  _^  7?  —  xy-\-\f-  _  {x  +  y)  {x^  —  xy-\-y^)  x  xy 
xy^      '  xy^  xy^  x  {x^  —  ^y  +  y^) 

=  ^  +  i^.    Ans,. 

y 

a{a  —  b)  +  b  (a^J))    ,    a  («  +  b)  —b{a  —  b) 
io. 


2 


{a  -b){a  +  b)        '         {a-  b)  {a  +  b) 

a2  ^  J2       a2  _  ^2 


14. 


—  fl2  _  j)i  ^  a?  j^  tt-i 
xy  +  2y^  +  x^  -^  xy  ^  a?  +  '2xy  +  2^/^ 

^(-^  +  y)         ^     y{x  +  y) 


=  1.     ^»s. 


y{x  +  y)  x^  +  2xy  +  2^ 

a*-l       x^  +  1        ix'^-l)ia^  +  l)(:^+  1)  ^^       a: 

^^'     '~'^^^^x~-                      a^  "^  (x^  +  1) 

—               a;3                                a:3  a;       a;^ 

a:4  _|.  ^2  +  1       a:2  _  a;  4-  l  _  x*  ^  x^  -\- 1       « 

1^'  ^  "^  ^  -  x2  ^a:2-a:  +  l 

_  .,^^+^2_+l    _                    _^   _^_^+  1        ^  1_     ^^^ 


68  KEY    TO    DAVIES'    BOURDON.  [81. 

17.  The  dividend  may  be  written  a-—{h-  +  2bc  +  c").  or  a^— (6  +  c)«, 
which  is  equal,  Prin.  3°,  Art.  33,  to  {a  +  b  +  c){a—b—e).  Hence,  the 
required  quotient  is  equal  to 

{a  -{-  b  -{-  c)  (a  —  b  —  c)       a  -\-  b  —  c 
i  a  +  ^"+~c 

=  [{a  -  t)  -  &]  X  [{a  -c)  +  b)]  z=  {a-cY-l^ 
=:  a^  —  2ac  -j-  0^  —  ^.     Ans. 


18. 


4. 


2fl2a;2  •       2ax 


—   X 


^4^_8ffV_-t-_12«4  _  a;2  —  2a^ 
ax  {x^  —  6a^)      ~       ax 


GENERAL    EXAMPLES. 


Ans, 


1+x^      1—x^  _  (1  +  2x'^  ^x^)  +  (I—  2x2  +  x^) 
1—  a;2"^l+2;2~  1  —  x* 

2  +  2a:4 

1  —X* 

..        1      + -^  =  (l^)±ii±f)  =  ^.    Ans. 

1  -\-  X         1  —  X  1  —  X^  1  —  3? 

a  +  b  _  a  —  b  _  (a^  +  2ab  +  b'')  —  (a^  —  2ab  +  ^>^) 
fl  —  i       a  +  *  ~  a2  _  j2 

4aJ  . 


a^  —  U^ 


1  +  3:8  _  ljZ_^  —  (1  +  2a:2  ^  ^4)  _  (i  _  2^2  +  a:^) 
1—  a;2       1+3^2""  1— a;4 

4a;2 
=  1^1^-     ^4^^- 


81-82.]  DIVISION    OF    FRACTIONS.  69 

^^     (a;-5)(.r-4)  ^  jx  -  6)  {x  -  7) 
z  {x  —  6)  x{x  —  5) 


a;2 
^-  (a:  +  J)  (:c  4-  J)  ><    (:r  -  6)    "  ^  ^^  +  *  ^ 

=  a;3  _|_  ^2-^,       ^^g. 

(7.) 

(a2  4-  2ax  +  rr^)  +  {a^—2ax  +  a:^) a?'' —  a^ 

«2_a;2  ^  (a2  +  2aa;  +  a;2)  — (a2_2aa:+a;«) 


^2(a2  +  a:2)  ^a^  +  a?      ^^^^ 
Aax  2ax 


g^     (7.+  l)  +  (^-l)    .   {n+l)-{n~l) 


9. 


10. 


w  +  1 

n  +  1 

2n         w  4-  1 

= X  — ^ —  =  n.     Ans. 

«  4-  1           2 

1   4.  a;2          1  _  a;2 

(1  4-  2a;2  4-  a^)  _  (1  _  2a;2  4-  x*) 

1  _  a;2      1  +  a;2  " 

1-x* 

4a;2 
=  ^_^.    Ans, 

1  —X         1  +x 

(l-:r)-(l+rr) 

1  —  a;2       1  —  a;2  " 

1  —  a;2 

2:?:            . 
\  —a? 

gi  4-  2g5  4-  ^2       g2  _  2fl5  4-  ^^  _  2  (g^  +  y^) 
"•  ^^ZTj         +        a2_52        -     «2_^,2    •     ^^«- 

14-^^       l_:,2_(]^^2)_(i^_     2a;2 


70  KEY    TO    DAVIES'    BOtJKDON.  [82-91. 

^„      1+^'       1  +  a:2  _  (1+^2)2 

(x  -b){x  +  b)  {x'  +  ^^)  ^-^     _  a:^+  ^ 

{x  —  b)  {x  —  b)  X  {x  +  b)  ~       X 

z=  X  A — •     Ans. 

X 
EQUATIONS   OF  THE   FIRST  DEGREE. 

.    ^.  bx      4x      ^„       7       132; 

1.  Given  ____13  =  --— . 

VERIFICATION. 

5  X  11.1  _  4  X  111  _  .0  _  Z  _  13  X  11.1 
13  3  8  G         • 

Multiply  by  24,  least  common  multiple, 

10  X  11 . 1  —  33  X  11 . 1  —  312  =  31  —  53  x  11.1;  that  is, 

—  556.2=  —556.2. 

2.  Given  a;   +  18  =  ox  —  ."),  to  linu  x. 
Transposing  and  reducing, 

—  2x=  —  23 ; 

dividing  both  members  by  —  2, 

x=  11^. 

3.  Given         6  -  2«  +  10  =  20  —  3a;  —  2,  to  find  «. 

Transposing  and  reducing, 

X  =  2. 

4.  Given  x  -\   -a;-|--a;=ll,to  find  x. 

Multiplying  both  members  by  6,  and  reducing, 

11a:  =  66; 
whence,  x  =  6. 


91.]  EQUATIONS    OF    THE    FIRST    DEOREE.  71 

5.  Given  2x — q*  +  1  =  5a:  —  2,  to  find  x. 

Multiplying  both  members  by  2,  transposing  and  reducing, 

-7a;  =  -6; 

whence,  j;  =  -. 

6.  Given  Sax  +  -    ~  3  =  bx  —  a,  to  find  x. 

Multiplying  by  2,  transposing  and  reducing, 

Gax  -2bx  =  6  —  Sa  ; 
factoring  the  first  member  of  the  equation,  we  have 

(6a  -2b)x  =  6-Sa; 
Q  —  3a 


whecce, 


X  z= 


Ga  —  2b 


7.  Given  — h  5  =  20 — ,  to  find  «. 

Multiplying  both  members  by  6, 

3a;  -  9  +  2x  =  120  —  3  a:  -|-  57  ; 
transposing  and  reducing, 

8a;  =  186  ;  .  • .  x  =  23|. 

i    r-                a;  +  3ar               ar  —  5 
3.  Given         — \-  -  =  4 — ,  to  find  x. 

/i  o  4 

Multiplying  both  members  by  12, 

6a;  -f  18  +  4a;  =  48  -  3a;  -(-  15  ; 
transposing  and  reducing, 

13a;  =  45  ;  • .  jr  =  3t^ 


I  ••' 


3  KEY    TO    DAVIES'    BOURDON.  [91. 


„    ^.              ax  —  b   ,    a       bx       bx  —  a         „    , 
9.  Given       — — -  +  -  =  — — ,  to  find  ar. 

Multiplying  both  members  by  12, 

3«x  —  36  +  4a  =  Ux  —  4bx  +  4a ; 
transposing,  reducing  and  factoring, 

3b 


(3a  —  2b)x  =  3b,  .  • .  x  = 


3a  —  26 


^    y^.  3aar       2bx        ^        y.       ^    -, 

10.  Given  • ; 4  =/,  to  find  x. 

c  a 

Multiplying  both  meiiibers  by  cd, 

Sadx  —  2bcx  —  Acd  ■=  fed ; 

transposing,  reducing  and  factoring, 

cdf  -}-  4it/ 
(3a^-26c)x^crf/+4crf;  .'.  ^  =  ^^^^T^; 

, .     /-  8aa;  ~  6       36  —  c  /.   .    c   j 

1 1 .  Given —  =  4  —  0,  to  find  x. 

7  2 

Multiplying  both'  members  by  14, 

16aa:  -  26  -  216  +  7c  =  56  -  146 ; 
cransposing  and  reducing, 

IGoiP  =r  56+ 96  — 7c;  .-.  x=-  ~    ' 


16a 


, ,    .^.  a;      a.  —  2  ,   a;       13        „   ^ 

12.  Given  -  —  _—  +  -  =  y ,  to  find  «. 

Multiplying  both  members  by  30, 

6a;  -  lOar  4-  20  +  15  ar  =  130 ; 
;ransposing  and  reducing, 

11a:  =  110;  '.  z  =  10. 


92.]  EQUATIONS    OF    THE    FIRST    DEGREE.  73 

X  X  X  X 

13.  Given  r  + j  =/  ^^  fi"<^  *• 

abed 

Multiplying  both  members  by  abed,  and  factoring, 

abcdf 
[bed  -  aed  +  abd  -  abe)  x  =abedf  .-.  x  =  ^^^  _^^^  ^  ^^^^  _-^- 


14.  Given     a; •—  H ^^  =  a;  +  1,  to  find  z. 

Multiplying  both  members  by  143, 

143x  -  33a;  +  55  +  52a:  -  26  =  143a;  +  143 ; 
transposing  and  reducing, 

19a;  =  114;  *.  x  ^  Q. 

15.  Given    ^  -  |-  -  ^^  =  -  l^ff.  to  «"<!  ^- 

Multiplying  both  members  by  315, 

45x  -  280a;  -  63x  H-  189  =  —  398S  ; 
transposing  and  reducing, 

-  298a;  =  -  4172  ;  .  • .  a;  =  14. 

-.^    n-  5(2a;  — 1)       {\—h)x,^, 

16.  Given     a; --^  =  ^^ t——  ,  to  find  x, 

0  +  1  0 

Clearing  of  fractions  and  performing  multiplications. 

h^x  -\-  bz  —  2i^x  +  W=.x  —  h^x; 
transposing,  reducing,  and  factoring, 

{J)  —  l)x=—h^',        .'.    x  =  Yzn' 


74  KEY    TO    DA  VIES'    BOURDOl!f.  [92'; 

17.  Given        3a;  -\ —  =  j  -j-  a,  to  find  x, 

o 

Multiplying  both  members  by  3, 

9x  +  bx  -  d  =  3x-{-Sa; 
transposing,  reducing  and  factoring, 

3a  +  d 


(6  +  b)x  =  3a  -\-  d;  .  • .  x  = 


6-J-6 


i«    r  ((/  +  b){x-  b)  4ah-b^  ,    a^-bz 

18.  Given  -^^ -^- —  Sa  = ; 2  x  +  - — ; ; 

a  —  b  a-\h  b        * 

to  tind  x. 

We  see  that  the  least  common  multiple  of  the  several 
Iractions  of  the  two  members  of  this  equation  is, 

Hence,  multiplying  both  members  of  the  equation  1^.' 
u~b  —  1^,  and  performing  all  the  indicated  operatiou.s,  v,  c 
shall  have, 

cfibx  4-  ^ah^x  +  b^x  —  aW  —  2ab^  —  ¥  —  3a^b  +  3ab^  =  ^nW 
—  hab^  +  ¥  —  la^bx  +  Wx  +  a*  —  a^bx  —  aW  +  i% ; 

then,  by  transposing, 

a^bx  +  'lab'^x  +  bH  +  'ia^x  —  Wx  +  a^bx  —  b^x  =  4:aW  —  baU 
+  J4  +  ^4  _  aW  +  aW  +  2ab^  +  b*  +  Sa^b  +  3ab^ ; 

factoring,  we  have, 

2b  (2a2  ^ab  —  b^)x  =  a*  +  3a%  +  4:aW  —  ^al^  +  2¥; 
dividing  by  the  coefficient  of  x, 

«*  +  3a^b  +  4a2J2  _  6aJ3  _|_  254 


a;  = 


2b  {2cfi  +  ab  —  b^) 


92. j  EQUATION'S    OF    THE    FIRST    DEGREE.  YD 

19.  Given,  x  =  3x  —  ^{4:  — x)  +  ^. 

Clearing  of  fractions,  and  di'opping  parenthesis, 

6x  =  18a:  —  13  +  Sx  +  2. 
Transposing  and  reducing, 

-  15a:  =  —  10 ; 
_2 

^.  ax  -\-  b      X  —  b^        1      J.    G   A  ^ 

20.  Given,     —^ 77-  =  3i »  to  fi°<l  ^• 

Multiplying  both  members  by  a%S 

aPx  +  b^  —  (fix  +  am  =  b^ 

Transposing,  factoring  and  reducing, 

«*' 
a{bf^-a)  =  -a  {ab^) ;  .:    x  =  ^^--r^- 

^,    ^.  2a: +  5       40  —  a:       10a:  -  427 

21.  Given,     _^^  + -^_  = —^^ . 

Clearing  of  fractions, 

304a:  +  760  +  9880  —  247a:  =  1040a:  —  44408. 
Transposing  and  reducing, 

—  983a;  =  —  55048 ; 
a:  =  56. 

„^^.  X       x  —  5       _  /2a:,    ^\ 

22.  Given, ^^  +  5=a:  —  (^—  +  lj. 

Clearing  of  fractions, 

11a;  —  7a;  +  35  +  385  =  77a;  -  2a:  —  77. 
Transposing  and  reducing, 

_  71.^  =  _  497 ; 

^^    r^-  x  —  \       x  —  2       X -\- 3       a:  +  4 

23.  Given,     ^—  +  -^—  =  —J-  +  -y-  +  1. 

Clearing  of  fractions, 

6a:  —  6  +  4a:  —  8  =  3a:  +  9  +  2.7;  +  8  +  12. 


76  KEY    TO    DAVIES'    BOURDON.  [92. 

Transposing  and  reducing, 

5a:  =  43 ; 

_.    ...            x  —  1       x  —  2       x  —  5       x  —  6 
24.  (jiven, r =  - — — -; . 

x  —  2       x  —  6       x  —  'o       x—7 

Performing  the  indicated  subtraction  in  botli  members, 
—  1  —  ] 


{x  —  2){x  —  3)       {X  —  6)  {X  —  1) ' 

Clearing  of  fractions,  and  performing  indicated  operations, 

:t2  _  13a;  +  42  =  x^  —  5x  +  (i. 
Transposing  and  reducing, 

—  Sx=  —36; 

25.  Given,  {x  +  ^a)  {x  —  ^b)  —  {x  +  a)(x  —  b)  -\- {b  =  0. 

Performing  multiplications  and  clearing  of  fractions, 

4.^•-  +  2ax  —  '2bx  —  ab  —  4a^  —  4aa;  -f-  '^bx  -\-  Aab  +  5  z=  0. 

Kcducing,  transposing:,  and  factoring, 

(3a  +  l)b 


—  2{a—b)x=  —  {Za-{-l)b',  .:     x 

26.   Given, 


2  (a  -  b) 

6.r  +  7       2a;  —  2       2a;  +  1 


15  7a;  —  6  5 

Clearing  of  fractions, 

42a;2  ^  13.^  _  42  —  30a;  +  30  =  42a;2  —  15a:  —  18. 
Transposing  and  reducing, 

—  2a;  =  —  6  ; 

•    a  X     O. 

1111 

27.  Given 


X  —  a       X  —  2a       X  —  3a       x  —  4a 
Performing  the  indicated  subtra  tions, 

—  a  —  a 


(x  —  a)  {x  —  2a)       {x  —  3a)  (x  —  4a) 


92.]  EQUATIONS    OF    THE    FIRST     DEGREE.  77 

Dividing  by  —  a  and  clearing  of  fractions, 

{x  —  3c5)  (a^  —  4fl)  =  {x  —  a){x  —  2a),    or 
xi  _  ^ax  +  Urr-  =  .7-2  —  3ax  +  Sa^. 
Transposing  and  reducing, 

—  iax  =  —lOa^ ;  .*.     x  =  |a. 

28.  Given,  {x  +  ly  =  {5  +  x)x  —  2. 

Performing  indicated  operations, 

x^  +  2a;  +  1  =  5:c  +  x'^  —  2. 
Transposing  and  reducing, 

_  3.?:  =  —  3 ; 

2  16 

Clearing  of  fractions, 

6a;2  —  20z  +  0  +  C^s  —  l'7x  +  5  =  12a;8  —  66a;  +  90. 
Transposing  and  reducing, 

29^^:79; 
_79 

„^    ^.  ic  a;  a 

30.  Given,  -  + 


Factoring, 
Reducing, 


a      b  —  a      b  -\-  a' 

i  1  1      j  _     a 

la       b  —  a  S       b  +  a' 

\a{b  —  a)f       {b  +  a)' 
a\b  -  a) 


X  = 


b{b  +  a) 
3i.0iven,    1(. -«)  -  ^(.  -  |)  +  !(.- |)  =  0. 


78  KEY    TO    DAVIES'    BOUEDON.  T  ,3-97 

Performing  indicated  operations, 

111111 

Clearing  of  fractions,  and  reducing, 

25a;  =  8a ; 
_8a 
''-25- 

32.  Given,       1  %x  -  '^^^  ~  •^-  =  Ax  +  8.9. 

.5 

Clearing  of  fractions, 

.6a;  —  .18:i-  +  .05  =  .2a;  +  4.45. 
Transposing,  and  reducing, 

.22a;  =  4.40 ; 
a;  =  20. 

33.  Given,       4.8a;  -  '"^^^  ~  '^^  =  1.6a;  +  8.9. 

.5 

Clearing  of  fractions, 

2.4a;  —  .72a;  +  .05  =  0.8a;  +  4.45. 
Transposing  and  reducing, 

.88a;  =  4.40 ; 


STATEMENT  AND  SOLUTION  OF  PROBLEMS. 

8.  Divide  $1000  between  A,  B,  and  C,  so  that  A  shall  have 
!:  *  I  more  than  B,  and  C  $100  more  than  A. 

liet        x        denote  the  number  of  dollars  in  B's  share. 
len  will  a;  +  72  "  «  «  "       A's     " 

;.  ,d  a:  +  72  +  100 "  «  «  «      C's      " 

From  the  conditions  of  the  problem, 


97.]  EQUATIONS    OF    THE     FIRST    DEGREE.  7'J 

^  _j-  ^  4.  72   f  a;  +  172  =  1000;     or,     3 a;  =  750  ,     .  • .    x  ^  252, 
or,  A's  share  is  $324,  B's  share  $252  and  C's  share  $424. 

9.  A  and  B  play  together  at  cards.  A  sits  down  v  ith  $84  and  B 
with  $48.  Each  loses  and  wins  in  turn,  when  it  appears  that  A  has 
five  times  as  much  as  B.     How  much  did  A  win? 

Let  X  denote  the  number  of  dollars  that  A  wins. 

Then  will  84  +  a;  denote  what  A  has  at  last, 

and  48  —  a;  what  B  has  at  last ; 

from  the  conditions  of  the  problem, 

84  +  a;  =  5  (48  —  x) ;         or,         84  +  ar  =  240  —  5a: ; 

whence,  .r  =  26 ;         or,         A  wins  $26. 

10.  A  person  dying,  leaves  half  of  his  property  to  his  wife,  one 
sixih  to  each  of  two  daughters,  one  twelfth  to  a  servant,  and  the 
Temaining  $600  to  the  poor  :   what  was  the  amount  of  his  property  1 

Let  X  denote  the  whole  number  of  dollars  in  the  property. 

Then  will   -       "  "  "  "         "  in  the  wife's  share. 


X 

6 


{( 


each  daughter's  '' 


a.,d  —       "  "  "  "         "  the  servant'? 

fiom  the  conditions  of  the  problem, 

multiplying  both  members  l\v  12,  transposnrg  and  reducing, 
~  X  =  —  7200  ;         or,         x  =  7-.200. 


80  KEY    10    DA  VIES'    BOURDON.  [97. 

11.  A  father  leaves  his  property,  amounting  to  $2520,  to  four 
sons,  A,  B,  C  and  D.  C  is  to  have  $360,  B  as  much  as  C  and  D 
together,  and  A  twice  as  much  as  B  less  $1000:  how  much  do  A. 
B  and  D  receive  1 

liet        X         denote  the  number  of  dollars  that  D  receives  • 
Then  will  x  -f  360       "  "  "  "    B 

and  2x  ^-  720  -  1000        "  "  "    A 

from  the  conditions  of  the  problem, 

360  -H  re  +  a;  +  360  +  2a:  -f-  720  -  1000  =  2520 ; 
transposing  and  reducing, 

4x  =  2080  ;  .  • .  a;  =  520 

or,     D's  share  is  $520  ;  B's  share  $880,  and  A's  share  $760. 

12.  An  estate  of  $7500  is  to  be  divided  between  a  widow,  two 
sons,  and  three  daughters,  so  that  each  son  shall  receive  twice  as 
much  as  each  daughter,  and  the  widow  herself  $500  more  than  all 
the  children :  what  was  her  share,  and  what  the  share  of  each  child  ? 

Let  X  denote  the  number  of  dollars  in  each  daughter's  share  ; 

Then  will  2x         "  "  "  «         son's  » 

and  4a;  -f-  3x  4-  500"  "  "         the  widow's  " 

from  the  conditions  of  the  problem, 

4a:  +  3a;  +  4a:  +  3a:  +  500  =  7500 ; 
transposing  and  reducing, 

14j?r=7000;  •.  a:  =  500. 

Daughters'  share  $500  ;  son's  share  $1000  ;   widow's  share  $400(». 

13.  A   company   of    180   persons   consists  of  men,    women   ari'i 


97-98.]  EQUATIONS    OF    THE    FIEST     DEGREE.  81 

children.  The  men  are  8  more  in  number  "^han  the  women,  and  the 
children  20  more  than  the  men  and  women  together  :  how  many  of 
each  sort  in  the  company  ? 

Let  X  denote  the  number  of  women; 
Then  will  a;  +  8    "  "  men  ; 

and  z  +  a;  +  8  +  20  "  children. 

From  the  conditions  of  the  problem, 

a:  +  x  +  8  +  a;  +  a;  +  8  +  20  =  180; 
transposing  and  reducing, 

4a;  =  144;  .  •.  x  =36. 

36  women,  44  men  and  100  children. 

14.  A  father  divides  $2000  among  five  sons,  so  that  each  elder 
should  receive  $40  more  than  his  next  younger  brother  :  what  is 
the  share  of  the  youngest  % 

Let  X  denote  the  number  of  dollars  in  the  youngest's  share. 

Then  will  x  -f-  40 "  "  "  "        second's 

a;  -1-  80 "  "  "  "        third's 

X  -\-  120  "  "  "        fourth's 

a:  +  160  "  "  "        fifth's 

From  the  conditions  of  the  problem, 

5a-  +  400  —  2000 ; 

transposing  and  reducing, 

5a;  =  1600 ;  .  • .  x^  320. 

15.  A  purse  of  $2850  is  to  be  divided  among  three  peric-us,  A, 
D  and  C ;  A's  share  is  to  be  y^y  of  B's  share,  and  C  is  to  have  $300 
more  than  A  and  B  together  :  what   s  each  one's  share  ? 


8JJ  KEY    TO    DAVIES'    BOURDON.  [98. 

Let         X  denote  the  number  of  dollars  in  B's  share. 

fir 

Then  will   ^     "  "  "  "      A's     " 

and  ^  +  n  "^  ^^^  "  "      ^'^     " 

From  the  conditions  of  the  problem, 

:^:  +  |^  +  :r  +  ^  + 300  =  2850; 

clearing  of  fractions,  transposing  and  reducing, 

34a;  =  28050  ;  .  • .  a.-  =  825  ; 

hence,  B's  $825  ;  A's  $450  ;  and  C's  $1575. 

16.  Two  pedestrians  start  from  the  same  point;  the  first  steps 
twice  as  far  as  the  second,  but  the  second  makes  five  steps  while  the 
first  makes  but  one.  At  the  end  of  a  certain  time  they  are  300  feet 
apart.  Now,  allowing  each  of  the  longer  paces  to  be  3  feet,  how  far 
will  each  have  travelled  ? 

Let  X       denote  the  number  of  feet  travelled  by  the  first. 

Then  will    \  "  "  steps 

£^  «  «  «  '-'■''      second. 

3 


5a;         15a; 
-3-'"'-6- 


aiid  4  X  ^,  or  ^  "  feet 


From  the  conditions  of  the  problem, 

i^  -  a-  =  300  ; 
6  ' 

clearing  of  fractions,  transposing  and  reducing, 


15a: 


9a;  =  1800 ;  . ' .  ir  =  200     ard      —  =  500. 


S8.]  EQUATIONS    OF    THE    FIEST    DEGREE.  83 

17.  Two  carpenters,  24  journeymen,  and  8  apprenticec,  received 
al  the  end  of  a  certain  time  $144.  Tiie  carpenters  received  $1  per 
day.  each  journeyman  half  a  dollar,  and  each  apprentice  25  cents.' 
bow  many  days  were  they  employed  1 

Let        X  denote  the  number  of  days. 

Then  will  x       "  "  dollars  due  each  carpenter. 


X 

2 


<(  u 


journeyman. 


and  7  u  i         apprentices ; 

from  the  conditions  of  the  problem, 

„     .   24x      Sx      ,,, 

reducing, 

16xz=144',  .-.  .r  =  9. 

18.  A  capitalist  receives  a  yearly  income  of  $2D40;  four  fiths  ol 
his  money  bears  an  interest  of  4  per  cent,  and  the  remainder  of  5 
per  cent. :  how  much  has  he  at  interest  1 

Let      X     denote  the  number  of  dollars  at  interest 

4a;         4 

Then  will  —  X  r^  denote  the  interest  of  1st  parcel. 

and  |x4       "  "  -^      " 

From  the  conditions  of  the  problem, 

4x         A         X         5         ^     ^ 

T  >^  loo  +  5  >^  100  =  «»**; 

i^learing  of  fractions,  and  reducing, 

21.C  =  1470000;  •  x  =  70000. 


84  KEY    TO    DAVIES'    BOFRDON".  [98. 

10.  A  cistern  containing  60  gallons  of  water  has  three  unequal 
cocks  for  discharging  it;  the  largest  will  empty  it  in  one  hour,  the 
second  in  two  hours,  and  the  third  in  three  :  in  what  time  will  tbt- 
cistern  be  emptied  if  they  all  run  together? 

Let  X  denote  the  required  number  of  minutes. 

Then  since   the  first  emits    1    gallon  per  minute,  the  second   \  of  a. 
gallon  per  minute,  and  the  third  g-  of  a  gallon,  , 

X  will  denote  the  number  of  gallons  emitted  by  the  1st. 


X 

2 


X 

3 


3d. 


From  the  conditions  of  the  problem, 


clearing  of  fractions  and  reducing. 

11a;  =  360  .-.  x  z=.2,%^^m. 

20.  In  a  certain  orchard  \  are  apple-trees,  \  peach-trees,  \  plum- 
trees,  120  cherry-trees,  and  80  pear-trees  :  how  many  trees  in  the 
orchard  % 

Let  X  denote  the  whole  number  of  trees. 

Then  will     -      "  "  "         "  apple-trees. 

-      "  "  "         "  peach-trees. 

^       "  "  "         "  plum-trees. 

From  the  conditions  of  the  problem, 


99.]  EQUATIONS    OF    THE    FIRST     DEGREE.  85 

clearing  of  fractions,  transposing  and  x'educing, 

—  x=-  2400  .  • .  a;  =  2400. 

21.  A  farmer  being  asked  how  many  sheep  he  had,  answered  that 
he  had  them  in  five  fields ;  in  the  1st  he  had  \,  in  tLa  2d  ^,  in  the 
od  i,  in  the  4th  ^,  and  in  the  5th  450  :  how  many  had  he  ? 

Let  X  denote  the  whole  number  of  sheep  : 

Then  will   %         "  "  "  "      in  1st  field. 


X 

6 

X 

8 


((  ((  ((  ((  iyA       u 

((  ((  (i  ((  f^j      i( 


and  —        "  "  "  "  4th    " 

12 

From  the  conditions  of  the  problem, 

4+6+8+  i2  +  ''^»  =  ^-' 
multiplying  both  members  by  24,  transposing  and  reducing, 

—  9a;  =  —  10800  .  • .  a;  =  1200. 

22.  My  horse  and  saddle  together  are  worth  $132,  and  the  horse 
is  worth  ten  times  as  much  as  the  saddle :  what  is  the  value  of  the 
horse? 

Let        X  denote  the  number  of  dollars  that  the  saddle  is  worth. 

1  hen  will  10a;     "  "  "  "       horse         " 

From  the  conditions  of  the  problem, 


86  KEY    TO    DAVIES'    BOUKDOJSf.  [99. 

a;  +  10a:=  132; 
reducing,     llx  =  132         .  •  .      a;  =  12;     whence,     10a;  =  120. 

23.  The  rent  of  an  estate  is  this  year  8  per  cent,  greater  than  it 
was  last.     This  )  ear  it  is  $1890  :  what  was  it  last  year  1 

Let        X  denote  the  number  of  dollars  in  last  year's  rent 
Then  will  ^  -^  ^  "  "  "  this     "         « 

From  the  conditions  of  the  problem, 

clearing  of  fractions  and  reducing, 

108a;  =  189000;  .  •,  x=  1750. 

24.  What  number  is  that  from  which,  if  5  be  subtracted,  f  of  ttif 
remainder  will  be  401 

Let  z  denote  the  number  required  : 

From  the  conditions  of  the  problem, 

I  (a; -5)  =40; 

clearing  of  fractions,  performing  operations   indicated,   transposing 
and  reducinf;^ 

2a;  =  130;  .  •.  a;  =  65. 

25.  A  post  is  ^  in  the  mud,  ^  in  the  water,  and  ten  feet  above  the 
water:  what  is  the  whole  length  of  the  post  ] 

Let  X  denote  the  number  of  feet  in  length. 

Then  will    ^         "  "  "  "  the  mud ; 

X 

and  M  tt  tt  M  jjjp  water : 


99.]  EQUATIONS    OF    THE     FIRST     DEGREE.  87 

From  the  ccnditions  of  the  problem  ; 

X  X 

-+-+10  =  a:; 

clearing  of  fractions,  transposing  and  reducing, 

—  52;  =  -  120 ;  .  • .  ar  =  24. 

20.  After  paying  i  and  ^  of  my  money,  I  had  66  guineas  left  c 
my  purse  :  how  many  guineas  were  in  it  at  first  % 

Let  X  denote  the  number  at  first ; 

from  the  conditions  of  the  problem, 

.-^--  =  66; 

clearing  effractions,  transposing  and  reducing, 

11a:  =  1320;  .-.  a;  =  120. 

t 

27.  A  person  was  desirous  of  giving  3  pence  apiece  to  some 
beggars,  but  found  he  had  not  money  enough  in  his  pocket  by  8 
pence ;  he  therefore  gave  them  each  two  pence  and  had  3  pence 
rem.aining  :  required  the  number  of  beggars. 

Let  X  denote  the  number  of  beggars ; 
then,  by  the  first  condition, 

3a;  —  8         denotes  the  number  of  pence  ; 
by  the  second  condition, 

2ar -f  3         denotes  the  lumber  of  pence  ; 
hence,  3a;  —  8  =  2a;  -|-  3  ; 

transposing  and  reducing 

a;  =11. 


88  KEY    TO    DAVIES'    BOURDOJST.  [99. 

28.  A  person  in  play  lost  -J-  of  his  money,  and  then  won  3 
shilling?  ;  after  which  he  lost  ^  of  what  he  then  had  ;  and  this  done, 
found  that  he  had  but  12  shillings  remaining:   what  had  he  at  first? 

Let  X        denote  the  number  of  shillings  at  first ; 

Then  will  ;r  -  ^  +  3       "  "  "       after  first  sitting , 

and  (:._^+ 3) -1(^-^  +  3) 

will  denote  what  he  finally  had  ; 

hence,  from  the  conditioiis  of  the  problem, 

^-i  +  3-i(.-^+3)  =  12; 

clearing  of  fractions,  performing  indicated  operations,   ttanspi.-'ing 
and  reducing, 

6a;  =  120  ;  .  • .  a;  =  20. 

29.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money  in  trade ; 
A  gains  $126,  and  B  loses  $87,  and  A's  money  is  now  double  B's : 
what  did  each  lay  out? 

J-^et  X  denote  the  number  of  dollars  laid  out  by  each  ; 

Then  will    x  +  126       "  «  "      A  had ; 

and  a;  —  87         "  "  "      B     " 

From  the  conditions  of  the  problem, 

X  +  126  =  2  (a;  -  87)  ; 
[jerforming  indicated  operations,  transposing  and  reducing, 
-  a;  =  -  300  ;  .  • .  X-  300. 

30.  A  person  goes  to  a  tavern  with  a  .ertain  sum  of  money  hi  his 


99-100.]         EQUATIONS    OF    THE    FIRST    DEGREE.  89 

pocket,  where  he  spends  2  shillings ;  he  then  borrows  as  much 
money  as  he  had  left,  and  going  to  another  tavern,  he  there 
spends  2  shillings  also;  then  borrowing  again  as  much  money  as 
was  left,  he  went  to  a  third  tavern,  where,  likewise,  he  spent  '^ 
shillings  and  borrowed  as  much  as  he  had  left ;  and  again  spend- 
ing 2  shillings  at  a  fourth  tavern,  he  then  had  nothing  remaining. 
What  had  he  at  first  ? 

Let  X  denote  the  number  of  shillings  at  first. 
Then,  from  the  first  condition, 

X  ~    2  will  denote  what  he  has  after  1st  visit. 

2{x  -    2)  -  2  or  2x  -    6         "  "  "         "    2d     '• 

2{2x—    G)— 2  or  4a;  — 14         "  "  '•         "    3d     " 

2(4x- 14)  —  2  or  8.C-30         "  "  "         "    4th     " 

From  the  conditions  of  the  problem, 

8.C  -  30  zz:  0     or     8a;  =  30 ;  .  • .  x  =  S^- 

or  the  amount  at  first  was  3*.  9af. 

31.  A  farmer  bought  a  basket  of  eggs,  and  offered  them  at  7  cents 
a  dozen.  But  before  he  had  sold  any,  5  dozen  were  broken  by  a 
careless  boy,  for  which  he  was  paid.  He  then  sold  the  remainder  at 
8  cents  a  dozen,  and  received  as  much  as  he  would  have  got  for  the 
whole  at  the  first  price.     Kow  many  eggs  had  he  in  his  basket  1 

Let        X  denote  the  number  of  dozens  at  first ; 

Then  will  a;  —  5       '  "  "        sold  ; 

And  8{x  —  5) "  "  cents  received ; 

Ix  '*  "  "     first  asked ; 

hence,  Ix  —  8  (x  —  5)  ;  .  • .  x  =  40. 


90  KEY    TO    DAVIES'    BOURDOX.  [100. 

32,  A  cask,  A,  contains  a  mixture  of  13  gallons  of  w^ne 
and  18  gallons  of  water ;  another  cask,  B,  contains  a  mixture 
of  9  gallons  of  wine  and  3  gallons  of  water :  how  many  gal- 
lons must  be  drawn  from  each  to  produce  a  mixture  of  7 
gallons  of  wine  and  7  gallons  of  water  ? 

Let  X  denote  the  number  of  gallons  drawn  from  the  cask 
A,  and  14  —  x  the  number  of  gallons  drawn  from  the  cask  B. 

Of  the  X  gallons  drawn  from  A,  ^ths  is  wine  and  ^ths  is 

water;    in  like  manner,  of  the  mixture  drawn  from  B,  ^ths 

is   wine,  and    j^ths   is   water.      Hence,   all   the   wine   drawn 

12  9 

from  both  is  equal  to  —a; +  -^(14  —  .r)  gallons;  but  this  is 

equal  to  7  gallons.     Hence, 

|.  +  l(U-,-)  =  7, 

or,  _:y  +  -(14  — ,r)=:7; 

a;  =  10     and     14  —  a;  =  4. 

33,  At  what  time  between  1  and  2  o'clock  is  the  minute 
hand  of  a  clock  just  1  minute  space  ahead  of  the  hour  hand? 

Let  X  denote  the  number  of  minute  spaces  passed  over  by 
the  hour  hand  fi'om  12  o'clock  till  the  hands  have  the  re- 
quired position.  Then  61  +  x  will  denote  the  number  of 
minute  spaces  passed  over  by  the  minute  hand  in  the  same 
time;  but  the  minute  hand  travels  12  times  as  fast  as  the 
hour  hand.     Hence, 

12a;  —  %\-\-x, 
or,  11a;  =  61 ; 

a;=5-j6i     and     61  +  a;  =  eO^Sj.. 

That  is,  the  hands  will  have  the  required  position  at 
Ih.  6^min. 


100.]  EQUATIONS    OF    THE    FIRST     DEGREE.  91 

34.  A  person  having  a  hours  at  his  disposal,  how  far  can 
he  ride  in  a  coacli  that  travels  b  miles  per  hour,  and  return 
on  foot  at  the  rate  of  c  miles  per  hour? 

Let  X  denote  the  number  of  miles. 

The  time  required  to  ride  x  miles  in  ihe  coach  vf'iW  be  de- 
noted by  'y,  and  the  time  required  to  walk  bi;ek  will  be  de- 

X 

noted  by  --.    From  the  conditions  of  the  problem,  we  have, 
c 

~  -\-  -  =za.    Hence, 
0        c 

yih  +  c)x  =  ahc;     .\    x  =  -^^~.j 

35.  A  can  do  a  piece  of  work  in  one-half  the  time  that  B 
can ;  and  B  can  do  it  in  two-thirds  the  time  that  C  can ; 
all  together  can  do  it  in  6  days.  How  many  days  would  it 
take  each  to  do  it  singly? 

Let  X  denote  the  number  that  it  will  take  A  to  do  it. 
Then  Avill  2x  denote  the  number  of  days  that  it  will  take  B 
to  do  it ;  and  3x  will  denote  the  number  of  days  it  will  take 

C  to  do  it.  Consequently,  A  can  do  a  part  denoted  by  - 
in  one  day,  B  can  do  a  part  denoted  by  — ,  C  can  do  a 
part  denoted  by   —   in   the   same  time,  and  all  together  can 

OX 

do  a  part  denoted  by   -  in  one  day.     Hence,  from  the  con- 


ditions  of  tht 

i  p 

roblem, 

1        1 

x~^  2x 

+ 

1 

3x~ 

1 
6' 

Factoring, 

^(- 

1 

2 

-l)  = 

1 

6' 

or, 

1 

X 

(6  +  3  +  2) 

=  1; 

1_ 

x~ 

1 
11 

,     or,     ; 

X  = 

=  11, 

2x 

=  22,    and 

3x  = 

33. 


92 


KEY    TO    DAVIES'    BOURDON. 


[106. 


SIMULTANEOUS   EQUATIONS   OF  THE  FIRST   DEGRER 

(  2x  +  3y  =  16  ) 
1.  Given  <  vto  find  x  and  y. 

Multiply  both  members  of  the  first  by  2,  and  of  the  second  by  3  ; 

4x  +  ijy  =  32  ) 
9^-  —  Oy  Z3  33  ) 

whence,  by  addition,  member  to  member,  we  have, 

13a;  =  65  ;  . ' .  a:  =  5,         also,         y  =  2. 


2.  Given 


(2x      3y_  9     ] 
y  "^  T  ~  20 

3a;       2tj  _  61 
I  T  "*"  y  ~  120  J 


.     to  find  X  and  v. 


Clearing  of  fractions,  and  then  multiplying  both  members  jf  the  first 
by  16,   and  of  the  second  by  5, 

128a:  +  240y  =  144) 
450ar  +  240y  =  305  ( 

whence,  by  subtracting,  member  from  member, 


322.r  =  161  ; 


3.  Given 


X  =  -,     also,  by  substit  ition,  v  =  — 
2  '    -^  '^3 


^+7^  =  99] 


I  +  7a;  =  51  I 


}■  to  find  X  and  y. 


Multiplying  the  first  by  343  and  the  second  by  7; 

49.1 

y 


).r   f  2401y  nr  33957  ) 
y   f    49a;     =     357     \ 


106-107.]       EQUATIOXS    OF    THE    FIRST    DEGREE.  93 

9y  subtraction, 
2400y  =  33000;     .-.     y  =  14  ;     also,  by  substitution,    a;  =  7. 


4.  Given 


f^       12-^4-8 


^^^.s^^^^l 


I" ;  to  find  ar  and  y 


Clearing  of  fractions  and  transposing, 

2a:  -  y  =  80 
47a:-  18y  =  2100; 

multiplying  both  members  of  the  first  by    18,  and   subtracting  the 
result  from  the  second,  member  from  member, 
11  a:  —  660 ;  .  ' .         a;  =  60  ;         by  substitution,         y  =  40. 


f).  Given  ' 


(x  +    y  +    2  =  29 
X  +  2//  -f-  32  =  62 

12^     3^    4 


(1) 
(2) 

(3) 


■  to  find  X,  y  and  z. 


Combining  (1)  and  (2), 

y  +  22  =  33     .     •     .     •     (4) ; 

combining  (1)  and  (3) 

2y  H-  32  =  54     •     •     •     •     (5) ; 
combining  (4)  and  f'5)  • 

2  =:  12;     by  successive  substitutions,    a-  =  8,     y  =  0. 

pa:  +  4y  -  32  =  22     •      •      (1)1 
6.  Given  -j  4.r  -  2y  +  52  =  18     •     •     (2)  j-;  to  find  r   /   and  2. 
[Qx  +  7y  -     2  =  63  (3)J 

Combining  (1)  and  (2). 


94 


KEY    TO    DAVIES     BOUKDON. 


[107. 


lOy- 

-112  =  26    . 

•                      • 

(4); 

eombining  (1)  and  (3), 

5y  —  82  =    3     . 

.     .     (5); 

combining  (4)  and  (5), 

5«  =  20  ; 

z 

By  successive  substitutions, 

X  =  3,         y 

r         y      z 

32 

7.  Given               < 

'?+»4-e  = 

15 

iv    ;   to  fi 

=  4. 

=  7. 


14       5       0 
Gearing  of  fractions, 

6a:  +    3y  +    22  =  192 

20a:  +  15y  +  122  =  900 

15a;  +  12y  4-  IO2  =  720 

combining  (1)  and  (2), 

16a:  +  3y  =  252     . 

combining  (1)  and  (3), 

15x  +  3y  =  240     .     .     . 

combining  (4)  and  (5), 

a:=^12; 

hy  successive  substitutions,     y  =  20,     z  =  30. 


(3); 


(4); 
(5); 


><.   Given  s 


(  7a;  —  22  +  3m  =  17 
4y  —  2z+      t=U 
5y  —  3a:  -  "Zu  =    8 
4y  —  3?^  +  2i!  =    9 
32  4-  Su  =  33 


(1)  ] 

(2) 

(3) 

(4) 

(5) 


;  to  find  X,  y, 
2,  t  and  u. 


EQUATIONS    OF    THE    FIRST    DEGREE. 


107.] 

Combining  (2)  and  (4), 

4y  —  4z  +  Su  =   13 

combining  (1)  and  (3), 

35y  -  63  -  5w  =  107 
combining  (5)  and  (6), 

12y  +  41m  =  171 
combining  (5)  and  (7), 

35y  -\-  Uu  =  173 

combining  (8)  and  (9), 

1303;/  =  3909  ; 


95 


•         •         • 


(6); 

(7); 
.  (8); 
.    (9); 

«  =  3; 


by  successive  substitutions,      x  =  2,     y  =  4,     z  =  3,     /=1, 


'  3a;  +  2?/  —  4z  =  15 
9.   Given,  i  5x  —  3y  +  2z  =  28 

dy  +  4z  —    x=2A 
Combining  (1)  and  (3), 

11^  +  8^  =  87    .    , 
Combining  (2)  and  (3), 

I2ij  +  22z  =  148  . 
Combining  (4)  and  (5), 

73^  =  365;    .-.    y  =  5. 
Substituting  and  reducing, 

a:  =  7,     and     2  =  4. 


(1) 
(2) 
(3) 

(4) 
(5) 


10.  Given, 


X  y 

1  1       ^ 

-+  -  =  2 

X  z 

1  1  _.3 

Ly  +  2-2 


(1) 
(2) 
(3) 


06 


KEY    TO    DAVIES     BOURDON". 


[108. 


Adding  (1),  (2),  and   (3),  member   to   member,  and   divid- 
ing by  2, 

1  1  1  u 

(4) 


1119 

-  -I 1 —  =  - 

X      y       z       4: 


Subtracting  (1),  (2),  and  (3),  successively,  from  (4), 


1_  5 

z~  4' 
1  _  1 
y~4: 
13 
x~  V 


z  = 


5 


X 


11.   Given, 


r  2      1  _3 

X     y~z 

z       y 

1       1  _4 

Vx^  2~  d 


Making    -  =  x', 


1         ,  .     1        , 

-  z=y ,     and     -  =  z , 

y     ^  2        ■ 

2x'  +    y'  =  dz'   . 
Sz'  —2y'  =  2      . 

.■+  .■  =  !    . 

Combining  (4)  and  (5), 

Ax' +  Sz' =  6z' +  2, 
or,  4:x'  —  Sz'  =  2      .     . 

Combining  (6)  and  (7), 

p, 

7x=6;    .■ 

Substituting  and  reducing, 

,       10 


X  —  ly ,     or,     X  —  _ . 


and, 


_  _21 

^  -21'    °^'    ^-10' 

y'=  -^,    or,    2/  =  - 


7 


(1) 
(3) 


(4) 
(5) 

(6) 


(7) 


108.] 


EQUATIONS    OF    THE    FIRST    DEGREE. 


07 


13.  Given, 


(di/  —  l_6z_x      9 

4       ~  5  ~3  "^  5 

5X  4rZ 

T  +  l 

dx-\-l 


T  +  y  =  ^  +  6 


14  +  6  ~  21  "^  3 


(1) 

(3) 


Clearing  of  fractions  and  transposing, 

loy  +  10a;  —  242  =  41 (4) 

—  12ij  +  15x  +  16^  =  10 (5) 

—  Uij-^18x—    '7z=  —13 (6) 

Combining  (4)  and  (5), 

115:c  —  162  =  214 (7) 

Combining  (4)  and  (6), 

410a;  —  441z  =  379 (8) 

Combining  (7)  and  (8), 

88312  =  8831 ; 
.-.      2  =  1. 
Substituting  and  reducing, 


X  =  2,      and      2/  =  3. 


a      b 


13.   Given, 


X  2   

a      c 

f +  7  =  1 
\  0        c 


Adding  (1),  (2),  and  (3),  and  dividing  by  2, 

a'^  h'^  c       2    '    ■    * 


(1) 
(3) 
(3) 


(^) 


98 


KEY    TO    DAVIES'    BOURDON. 


[108. 


Subtracting  (1),  (2),  and  (3),  successively,  from  (4), 


14.  Given, 


z 
c 

1 

2'    •'• 

c 

y  - 

1 

b 

y=r 

X 

a  ~ 

1 

2'    •  • 

a 

'    ^x- 

-37/=rl 

\\z  - 

-lu  =  l 

Az  - 

-  7.y  =  1 

19a;- 

-  3%  =  1 

Combining  (1)  and"  (4), 

21?<  —  57?/ =  12  .    .    . 
Combining  (2)  and  (3), 

28m -772/ =  7    •    •    • 
Combining  (5)  and  (6), 

32/  =  27;     .'.    i/  =  9. 
Substituting  and  reducing, 

a:  =  4,     z  =  16,     and     u  =  25. 
(dx-by  2x±y^ 


15.   Given, 


4       ~2  "^  3 


Clearing  of  fractions  and  transposing, 

l\x  —  my  =  —  30 
9a;  —    2?/  =  96     . 
,»,  y  =  6,    and     a;  =  12, 


(1) 
(3) 
(3) 
(4) 

(5) 

(6) 


.  .  (1) 

.  .  (2) 

.  .  (3) 

.  .  (4) 


108-111.]      EQUATIONS    OF    THE    FIRST    DEGREE.  99 

.      .       (1) 


r  3x       y        4  x        y 

I  T0~15~  9  ~l2  ~18 
16.   Given, 


Clearing  of  fractions,  transposing,  and  reducing, 

.39.r  -  3y  =  80 (3) 

115a:  H-  %  =  236     . (4) 

Combining  (3)  and  (4), 

193a;  =  386 ; 
a:  =  3 ,    and    y  =  — 1. 

PROBLEMS  GIVING   RISE   TO   SIMULTAxNEOUS   EQUATIONS  OP 

THE   FIRST   DEGREE. 

I 

5.  "What  two  numbers  are  they,   whose  sum   is  33  and  whose 
iifference  is  7  1 

Let  X  denote  the  first,  and  y  the  second. 
From  the  conditions, 

X  +  y  =  33 
ar-y=    7; 
whence,  by  combination, 

x  =  20.  y  =  13. 

().   Divide  the  number  75  into  two  such  parts,  that  three  time.s  ^hp 
greater  may  exceed  seven  times  the- less  by  15. 

Let  X  der.ote  the  greater,  and  y  the  less 
From  the  conditions  of  the  problem. 


100  KEY    TO    DAVIES'    BOUKDON".  [111-112. 

a;  +    y  =  75 

3^  -  7y  =  15  ; 

by  combination,     lOy  =  210 ;         .-.     y  =  21  ;     also,     x  =  54. 

7.  In  a  mixture  of  wine  and  cider,  \  of  the  whole  plus  25  gallons 
was  wine,  and  \  part  minus  5  gallons,  was  cider  :  how  many  gallons 
were  there  of  each  % 

Let         X  denote  the  number  of  gallons  of  wine  ; 

and  y         "  "  "  cider; 

"  mixture. 

+  25  =x 

3    -  5=y; 

clearing  of  fractions,  transposing  and  reducing, 

y  —  X  z=  —  50 

—  2y  +  a:  =        15 
by  combination, 

y  =  85  ;         and         a;  =  85 

8.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  and  tnf 
number  of  pieces  of  both  >orts  that  were  used  was  just  100;  if  the 
guineas  were  estimated  at  21s.,  and  the  moidores  at  27s.,  how  many 
were  there  of  each  ? 

Let     X  denote  the  number  of  moidores ; 

and         y         "  "  "  guineas ; 

then,  sitce  X120  =  2400s.,  we  have,  from  the  conditions, 


then  will    x  -\-  y 

(( 

From  the  conditions. 

x  +  y 
2 

x+  y 

112.] 


101 


EQUATIONS    OF    THE    FIRST    DEGREE. 

X  +       y  =     100 
27x  +  21y  =  2400  ; 
by  combination,     6y  =  300 ;     .  ' .     3/  =  50 ;     also,     x=  60. 

9.  Two  travellers  set  out  at  the  same  time  from  London  and 
York,  whose  distance  apart  is  150  mile*  ;  they  travel  toward  each 
other;  one  of  them  goes  8  miles  a  day,  and  the  other  7;  in  what 
time  will  they  meet? 

Let  X  denote  the  number  of  miles  travelled  by  the  first; 

"        second ; 

"       first ; 


y 

X 


i( 


then  will     - 
o 


(( 


days 


7 


and 

From  the  conditions. 


M 


tt 


u 


second ; 


a;  +  y  =  150; 


X       y 

8  ~  7  *' 
whence,  by  combination, 

X 

ar  =  80         and         -  =  10,         the  number  of  days. 

10.  At  a  certain  election,  375  persons  voted  for  two  candidates, 
and  the  candidate  chosen  had  a  majority  of  91 ;  how  many  voted  for 
each? 

Let  X  denote  the  number  of  votes  received  by  the  first; 
y        "  "  "  "  "       second; 

from  the  conditions  of  the  problem, 

a;  +  y  =  375 
x=z  y  +    91 ; 
by  combination,  ^  x  ■=  233,         y  =  142. 


102  KEY    TO    DAVIES'    BOUKDON.  [112. 

11.  A's  age  is  double  B's,  and  B's  is  triple  C's,  and  the  sum  of 
all  their  ages  is  140  :  what  is  the  age  of  each  1 

Let  .r  denote  the  age  of  A  ; 

?/  "  "  B; 

2         "  "  C; 

from  the  conditions  of  the  problem, 

X  =    2y     '     '     '     '     (1) 

y  =    Sz     ■     '     '     •     (2) 

x-\-y+  z=zUO     .     .     .     •     (3) 

from     (1)  and  (2),  a:  =  62  ; 

substituting,  y  =  Sz,  and     x  =  6z,    in  (3),  and  reducing, 

lOz  =  140  ;  .  • .  z  =  U,     X  =  84,     y  =  42. 

12.  A  person  bought  a  chaise,  horse  and  harness,  for  £60;  the 
norse  came  to  twice  the  price  of  the  harness,  and  the  chaise  to  twice 
the  price  of  the  horse  and  harness:  what  did  he  give  for  each? 

Let  X  denote  the  number  of  pounds  paid  for  the  harness; 

y         "  "  "  "         "       horse; 

z         "  "  "  "         "       chaise; 

from  the  conditions  of  the  problem, 

y  =  2x     .     •     .     .     (1) 

z  =  2{x+  y)     '     .     .     .     (2) 
a;  +  y  +    z  =z  GO     •     •     •     •     (3) 
from  (2)     and     (1)     2  =  6x; 

substituting         2  =  6x      and      y  =  2x      in     (3) 
9x  =  60,     .  • .    X  =  62,     also,  by  substitution,    y  =  13^,     z  =  40  ; 

hence,  the  price  of  the  chaise  was  £40;  of  the  horse  £13  Cts.  8c/.  • 
and  that  of  the  harr-iss  £6  13a-.  4d. 


112.]  EQUATIONS    OF    THE    FIRST    DEGREE.  103 

13.  A  person  has  two  horses,  and  a  saddle  worth  £50 ;  now,  if 
the  saddle  be  put  on  the  back  of  the  first  horse,  it  will  make  his 
value  double  that  of  the  second  ;  but  if  it  be  put  on  the  back  of  the 
second,  it  will  make  his  value  triple  that  of  the  first :  what  is  the 
value  of  each  horse  1 

Let     X     denote  the  number  of  pounds  the  1st  horse  is  worth; 
y         «  "  <  "     2d       "  '• 

from  the  conditions  of  the  problem, 

a;  +  50  z=  2y 

1/  -{-  50  =  Sx; 
whence,  by  combination, 

X  =     30     y  =    40. 

14.  Two  persons,  A  and  B,  have  each  the  same  income.  A  saves 
^  of  his  yearly ;  but  B,  by  spending  £50  per  annum  more  than  A. 
at  the  end  of  4  years  finds  himself  £100  in  debt ;  what  is  the  income 
of  each  ? 

Let     X     denote  the  number  of  pounds  in  the  income  of  A  ; 
y         «  "  "  "         "  B; 

by  the  conditions  of  the  problem,  these  are  equal ;  one  only  will  be 

used.     Then  will 

4 

-X     denote  what  A  spends  per  year : 

fx  -i-  50  "       "    B       " 
5 

from  the  conditions  of  the  problem, 

4^  +  50  j  =4.r  +  100; 

whence,  performing  indicated  operations,  transposing  and  reducing, 
4.r  =    500  .  • .  X  =     125. 


104  KEY    TO    DAVIES'    BOURCON".  [113. 

15.  To  divide  the  number  36  into  three  such  parts,  that  |  of  the 
first,  ^  of  the  second,  and  ^  of  the  third,  may  be  all  equal  to  each 
other. 

Let  a-,  y    and  2,  denote  the  parts. 
From  the  conditions  of  the  problem, 

X  -\-  y  -^  X  —  Z<S 

2~3 

X       z 
2~4' 

clearing  of  fractions,  and  combining, 

9z  =  72     .  • .     a;  =  8  ;     whence,     y  =  12     and     z  =  IH. 

16.  A  footman  agreed  to  serve  his  master  for  £8  a  year  and 
livery,  but  was  turned  away  at  the  end  of  7  months,  and  received 
only  £2  13s.  4c/.  and  his  livery  :  what  was  its  value? 

Let  X  denote  the  value  of  livery,  expressed  in  shillings :  £8  = 
160s.,  and  £2  13s.  4c7.  =  53is.  ; 

Then  will  ( — — — I  denote  the  value  of  wages  1  month, 

by  the  conditions  of  the  problem, 

/IfiO  +  a-\ 

1120  + 7a;  =  640  +  12  ^ 
—  hx=—  480 
ar  =  96      *.     value,  £4. 16s. 


113.]  EQX'ATIOXS    OF    THE    FIRST    DEGREE.  105 

17.  To  divide  the  numl)er  90  into  four  such  parts,  that  if  the  first 
be  increased  by  2,  the  second  diminished  by  2,  the  third  multiplied 
by  2,  and  the  fourth  divided  by  2,  the  sum,  difference,  product 
and  quotient,  so  obtained,  will  be  all  equal  to  each  other. 

Let     X,  y,  z    and  u,  denote  the  parts; 
from  the  conditions  of  the  problem, 

x+^=y -^ 
X  +  2  =  2z 

whence  we  find  from  the  last  three  equations, 

X 

y  =  X  +  4:,     z  =  -+l,     and     «  =  2a;  +  4 ; 

substituting  these  values  in  the  first  equation, 

.r  4-  a;  +  4  +  ^  4-  1  +  2a;  +  4  =  90  ;  or  4|a;  =  81  ;  .  • .    jr  =  18 ; 
whence,  by  substitution,     y  =  22,     2  =  10,     and     u  =  40. 

18.  The  hour  and  minute  hands  of  a  clock  are  exactly  together  at 
12  o'clock  :  when  are  thev  next  together. 

\st  Solution. 
Let  X  denote  the  number  of  minute  spaces  passed  by  the  hour 
hand  before  they  come  together ; 

and  y  the  number  passed  by  the  minute  hand  ; 

then,   since  the  latter  travels   12  times  as  fast  as  the  former,  and 

since  it  has  to  gain  60  spaces,  we  have, 

X  —  ?/  =  60 

x=l2y\ 


106  KEY    TO     DA  VIES'    BOURDON.  [113. 

bv  combination, 

lly  =  60  .  • .  y  —  5^Y  ;      also,      x  =  65^  ; 

hence,  they  will  be  together,  65-^  minutes  after  12  o'clock,  or  al    I 

o'clock,    -j^    minutes,   and   at   the  end   of  every  succeeding    eqiuil 
portion  of  time. 

2c?  Solution. 

The  minute  hand  will  pass  the  hour  hand  11  times  before  they 
again  come  together  at  12  o'clock,  and  the  times  between  any  two 
consecutive  coincidences  will  be  equal.  Hence  each  time  will  be 
equal  to  12  hours  divided  by   11  =  l-^hr.  =  Ihr.  5-^m. 

19.  A  man  and  his  wife  usually  drank  out  a  cask  of  beer  in  12 
days ;  but  when  the  man  was  from  home,  it  lasted  the  woman  30 
days ;  how  many  days  would  the  man  be  in  drinking  it  alone  ? 

Let     X     denote  the  number  of  days  it  takes  the  man  to  drink  it ; 

y         "  "  "  "  woman  "         " 

then,  if  the  whole  quantity  of  beer  be  denoted  by  1, 

-  will  denote  the  quantity  drank  by  the  man  in  1  day  ;     and 


woman  ; 


y 

from  the  conditions  of  the  problem, 

x^  y       12 
1  _  J_ 

y~30' 

substituting  the  value  of  -  in  the  first  equation, 

1       J__  J_ 
z  "^^  30  ~  12  "' 


113.]  EQUATIOXS    OF    THE    FIEST    DEGREE.  107 

clearing  of  fractions, 

QO  +  2x  =  5x',  .  • .  X  =  20. 

20.  If  A  and  B  together  can  perform  a  piece  of  work  in  8  days 
A  and  C  together  in  9  days,  and  B  and  C  in  10  days  :  how  many 
days  would  it  take  each  person  to  perform  the  same  work  alone  1 

Let  the  work  be  denoted  by   1  ; 

Let     X     denote  the  work  done  by  A  in  one  day  ; 

y  a  Ci  "  g      u  u 

2  "  <(  (i  p       (i  (' 

then  will  -,  -    and  -  respectively  denote  the  number  of  days  that 

it  will  take  A,  B,  and  C  severally  to  do  the  work  ; 

from  the  conditions  of  the  problem, 

x  +  y=    I     ....      (1) 
x-{-  z—    \     .     .     .     .     (2) 

y  +  ^-A    •    •    •    •    (3); 

clearing  of  fractions, 

8.t;  +    Sy  =  1     .     .     .     .     (4) 

9.C  +    9^  =  1     .     .     .     .     (5) 

10y+10s  =  l     .     .     .     .     (6); 

combining     (4)     and     (5), 

72y  -  722  =  1     •     .     .     .     (7), 
combining     (6)     and     (7), 

1440y  =  82  .-.  y  =  ^\ 

substituting  in     (1)     and     (3), 


a;   —  1 4J_   _      49     .  -r   _   _1 4 


8  7  20    —    720'  "    —    10  TSO 


31     . 

yao  » 


l  =  14fA;  ^=17|1;        -^  =  23^^. 


108  KEY    TO     DAVIES'    BOURDON".  [113-114. 

21.  A  laborer  can  do  a  certain  work  expressed  by  a,  in  a  time 
expressed  hy  b  ;  a  second  laborer,  the  work  c  in  a  time  d;  a  third, 
the  work  e  in  a  time/.  Required  the  time  it  would  tai\e  the  three 
laborers,  working  together,  to  perform  the  work  ff  1 

If  a  laborer  can  do  a  piece  of  work  denoted  b}  o,  in  a  number  of 
days   denoted   by  b,  he  can  do  in  1  day  so  much  of  the  work   as  is 

denoted  by  -r ;  the  second  in  1  day  can  do  so  much  as  is  denoted  by 

c  ^ 

-:  and  the  third  so  much   as   is  denoted  by-:    hence,    the    three 
d  J  ^  >  ) 

working  together  can  do 

a       c        e  _  ad/  -j-  bcf  +  bde 
b       ~d^f~  bdf 

Let  X  denote  the  time  required  to  perform  the  work  ^;    then, 
the  three  can  perfoi-m  the  work  -  in  the  time  1 ; 

from  the  conditions  of  the  problem, 

(J       adf  +  bcf  +  bde 
x~  bdf 

taking  the  reciprocals  of  each  member,  and  then  clearing  effractions. 

we  have, 

^^ bdfy 

ad/+  bcf+  bde 

In  this  example  only  a  single  unknown  quantity  has  been  used, 
and  it  may  be  remarked  that  many  other  examples,  in  this  chapter, 
may  be  more  easily  solved  by  a  single  unknown  quantity  ;  in  such 
cases  more  than  one  has  been  used  for  the  purpose  of  illustration. 

22.  If  32  pounds  of  sea  water  contain  1  pound  of  salt,  how  much 
fresh  water  must  be  added  to  these  32  pounds,  in   order  that  the 


114.1  EQUATIONS    OF    THE    FIRST    DEGREE.  109 

qiiaiiUty  of  salt  contained  in  32  pounds  of  the  new  mixture  shall  be 
reduced  to  2  ounces,  or  A  of  a  pound  1 

Let  z  denote  the  number  of  pounds  to  be  added  ; 

then  will denote   the  number  of  pounds   of  salt    in  each 

32  +  ^ 

pound  of  the  mixture,  but  this  we  know  to  be  -—  X  -,  or,  -— -. ; 

o2       o  2ob 

hence   from  the  conditions  of  the  problem, 

i =  — ,     or,     S2-{-  X  =  250,     or.     x  -  224. 

32  +  x       256'        '  '         ' 

This  problem  is  also  solved  by  a  single  unknown  quantity  more 

readily  than  by  two. 

23.  A  number  is  expressed  by  three  figures  ;  the  sum  of  theso 
figures  is  11  ;  the  figure  in  the  place  of  units  is  double  that  in  the 
place  of  hundreds  ;  and  when  297  is  added  to  this  number,  the  sum 
obtained  is  expressed  by  the  figures  of  this  number  reversed. 
What  is  the  number? 


> 


Let  X,  y    and  z  denote  the  digits  in  their  order : 
then  will  the  number  be  denoted  by 

lOOx+lOy+z; 
from  the  conditions  of  the  problem, 

x+       y^z  =  \\ (1) 

z  =    1x (2) 

100*  +  lOy  +  2  +  297  =  1002  +  lOy  +  a;     •     •     .     (3); 

reducing  (3),  gives 

992  -  99x  =i297 (1^; 

substituting  ?  =  2a;  in  (4),  and  reducing, 

99.r  =  297  ;  .  * .  x  =  3  ; 


110  KEY    TO    DAVIES'    BOURDOX.  [114. 

wht'nce,  l)v  successive  substitutions, 

y  =  2,         2  =  6.  Ans.  326. 

24.  A  person  who  possessed  $100000  dollars,  placed  the  greatei 
part  of  it  out  at  5  per  cent,  interest,  and  the  other  part  at  4  pe> 
cent.  The  interest  which  he  received  for  the  whole  amounted  U 
4640  dollars.     Required  the  two  parts. 

Let     X     denote  the  greater  part ; 
y         "         "   lesser       " 
From  the  conditions  of  the  problem, 

5ar 

— —  =  interest  on  x  dollars  at  5  per  cent. ; 

then. 


^            "         1/       " 
100                     -^ 

4     " 

x-\-y  =  100000     .     . 

•     •     (1) 

100+100-^^^^   •   • 

•     •     (2) 

clearing  (2)  of  fractions, 

5x  +  4p  =  464000     •     •     .     .     (3) 
combining  (1)     and     (3), 

y  =  36000,         whence         x  =  64000. 

25.  A  person  possessed  a  certain  capital,  which  he  placed  out  at 
a  certain  interest.  Another  person  possessed  10000  dollars  more 
than  the  first,  and  putting  out  his  capital  1  per  cent,  more  advan- 
tageously, had  an  income  greater  by  800  dollars.  A  third,  possessed 
1.5000  dollars  more  than  the  first,  and  putting  out  his  capital  2  per 
f'cnt.  more  advantageously,  had  an  income  greater  by  1500  dollars. 
Required  the  capitals  and  the  three  rates  of  interest. 


114.]  EQUATIONS    OF    THE    FIRST    DEGREE.  Ill 

Let  X  denote  the  number  of  dollars  in  1st  capital ; 
and       y  the  rate  per  cent. ;  then, 


(^  X  y) 
100 


will  denote  the  number  of  dollars  of  1st  income; 


\x  -f-  10000)  (i(  -^   1)  „  ,.  «         .,    2d 

100       " 


(.c  f  15000)  (.y  4-  2) 
100 

from  tbo  conditions  of  the  problem, 


"   3d        " 


{x  4-  10000)  (y  +  1)       ^  X  y 

^100" =  Too"  ■*"  ^^^ 

(;r+15000)(y  +  2)  _  ^^ij/   ,    1 500  • 
100  ~    100    "^ 

clra/'ng  of  fractions,  performing  indicated  operations,   transposing 
ar/]  reducing, 

lOOOOy  -{-    x=    70000 

15000y  + 2z  =  120000; 
c'ombininof  and  reducing, 

5000y  =  20000         .  • .         y  =  4 ;  and 

Dy  substitution,  x  =  30000 ; 

2d.   $40000,     rate  5  per  cent. 
3d.  $45000,     rate  G     "       " 

26.  A  cistern  may  be  filled  by  three  pipes,  A,  B,  C.  By  the  two 
first  it  can  be  filled  in  70  minutes;  by  the  first  and  third  it  can  be 
nlled  in  84  minutes;  and  by  the  second  and  third  in  140  minutes. 
What  time  will  each  pipe  take  to  do  it  in?  What  time  will  be 
required,  if  the  three  pipes  run  together? 

Call  the  contents  of  the  cistern  1. 


113  KEY    TO    DA  VIES'    BOURDON.  [115. 

Let  X  denote  the  quantity  discharged  in  1  minute  by  the  first ; 
y         «  '  «  "         "         "        second ; 

g         «  «  «  «         "         "        third ; 

then  will    -      -      and     -     denote  the  number  of  minutes  required 
X      y  z 

fur  the  pipes,  separately,  to  fill  the  cistern  ;         and, 

1 

a;  +  y  +  2' 

will  denote  the  number  of  minutes  required  for  all  three  tv    fill  it, 

running  together ; 

from  the  conditions  of  the  problem, 

^"^^~70     *     *     '     '     ^^ 
2  +  2  =  g^     •     •     •     •     (2) 

y  +  ^  =  lio  '  *  *  *   ^^^ ' 

clearing  of  fractions, 

70a:  +    ^Oy  =  1  •  •  •  •  (4) 

84a;  +    84z  =  1  •  •  •  •  (5) 

HOy  +  1402  =  1  •  •  •  •  (6) ; 

combining   (1)  and  (2), 

840?/  —  8402  z=  2  •  •  •  •  (7)  ; 

combining  (6)   and  (7), 

1 
ir>80y  =  8,  .-.  2/ =  210* 

substituting  in  (1),  and  transposing, 

_  ^  __L_  A  -_L. 

'  "to"  210  ~  210  ~  105' 


115.]  EQUATIONS    OF    THE    FIKST    DEGKEE.  113 

substituting  in     (3), 

_  J 1^  _  J_ 

^  ""  140  ~  2l0  ~  420' 

1  1,1  1 

X  -i-  y  +  z  =^777;  +  TTTT  4- 


210  '    105  '   420       60' 

hence,     -  =  105.      -  =  210,      -  =  420.    —, —  =  GO. 

X  '      y  z  X  -{-  y  -\-  z 

27.  A  has  3  purses,  each  containing  a  certain  sum  of  money.  If 
$20  be  taken  out  of  the  first  and  put  into  the  second,  it  will  contain 
four  times  as  much  as  remains  in  the  first.  If  ^60  be  taken  from  the 
second  and  put  into  the  third,  then  this  will  contain  If  times  as 
much  as  there  remains  in  the  second.  Again,  if  $40  be  taken  from 
the  third  and  put  into  the  first,  then  the  third  will  contain  2f  times 
as  much  as  the  first.     What  were  the  contents  of  each  purse  1 

Let     X     denote  the  number  of  dollars  in  the  first  purse. 
y         «  "  "  *'      second  " 

z         "  "  "  "      third      " 

then  from  the  conditions  of  the  problem, 
4  {x  —  20)  =y  -\-  20, 
l{y-  60)  -_.  2  +  60, 
2-40    ^  ^  (x  +  40)  ; 

clearing  of  fractions,  performing  operations  and  transposing, 

4x  -      y  =     100  ....     .  (1) 

7y  _    4z  =    660  .     .  (2) 

82  -  23a:  =  1240  ...  (3) ; 
combining     (1)     and     (2), 

28a;  -  4z  z=  1360  ....  (4) ; 


114  KEY    TO    DAVIES'    BOURDON.  [115. 

combining     (3)     and     (4), 

33a?  =  3960;     .'.     a;  =  120  ; 
Dy  substitution,  y  =  380;     z  =  500. 

28.  A  banker  has  two  kinds  of  money ;  it  takes  a  pieces  of  the 
first  to  make  a  crown,  and  b  of  the  second  to  make  the  same  sum. 
Some  one  offers  him  a  crown  for  c  pieces.  How  many  of  each 
kind  must  the  banker  give  him  ? 

Since  it  takes  a  pieces  of  the  first  to  make  1  crown,  -  —  the  part 
of  a  crown  in  each  piece ;  and   -,    the  part  of  a  crown   in   each 

piece  of  the  second: 

let     X     denote  the  number  of  pieces  taken  of  the  first  kind, 
y  «  «  «  "  second  " 

from  the  conditions  of  the  problem, 

X  -{-  y  —c 

-  +  f-  =  1,     or     6x  +  ay  =  a6  ; 
a       0 

by  combination, 

by  —  ay  z=  he  —  ab\     or     y  (6  —  a)  =  6  (c  —  o) ; 

(a  —  c)b  ,  a(c  —b) 

.-.     y  =  - r^-;      whence,     x  z= .-• 

^        a  —  b  a  —  b 

2!).  Find  what  each  of  three  persons,  A,  B,  C,  is  worth,  knowing, 
1st,  that  what  A  is  worth  added  to  I  times  what  B  and  C  are  worth, 
is  equal  to  p ;  2d,  that  what  B  is  worth  added  to  yn  times  what  A 
and  C  are  worth,  is  equal  to  q;  3d,  that  what  C  is  worth  added  to 
n  times  what  A  and  B  are  worth,  is  equal  to  r. 


115.]  EQUATIONS    OF    THE    FIRST    DEGREE.  IIS 

Let     I     denote  what  A  is  worth, 
y         «•  "       B 

2  "  "         C  " 

toen,  from  the  expressed  conditions. 

x-\-l  {y  +  z)  =p  '  •  •  .  (1) 
y  -\- m  {x  -\-  z)  =  q  •  •  •  •  (2) 
z  ■{-  n  {x  +  y)  =  r     •     •     •     •     (3)  ; 

wfiich,  by  adding  and  subtracting  Zx,  my    and  nz,  may  be  written 
under  the  forms 

(\-  l)x-\-  l{x^y  +  z)  =  p  .  .  .  .  (4) 
{\ —m)y +  m{x  +  y -\- z)  —q  •  -  •  .  (5) 
(I  —  n)  z  -\-  n  (x  -{-  y  -\-  z)  —  r     •     •     •     •      (6)  ; 

dividing  both  members  of  each  equation  by  the  co-efficient  of  its 
first  term, 

I 


X+  ,^—-{x  +  y  +  z)=   ^^      ....     (7) 

y  +  -, {x-\-y+  z)  =  -^—      ....     (8) 

z  4-  :j {x  +  y  -\-z)  =   •     •     •     •     (9)  ; 


adding  these,  member  to  member,  and  deducing  from  the  resulting 
equation  the  value  of  a:  +  y  +  2, 


1  —  I       \  —  m       1  —  n 
.  +  y  +  .  =  - ; —     ■     ■     (10). 

1  —  I       I  —  m       1  —  » 

Denote  the  second  merabei  of  equation   OO)   by  the  single  letter  s. 


116 


KEY    TO    DAVIES'    BOURDON. 


[115. 


a  known  quantity.  Then  by  substituting  this  for  the  factor 
j^  f  y  4-  2,  in  each  of  the  equations  (7),  (8)  and  (9),  and  deduo- 
'•g  the  values  of  x,  y    and  z,  we  have, 


X  — 


P 


Is 


p  —  Is 


z  = 


1 

■  I 

1 

— 

I 

1 

— 

■I 

9 

— 

ms 

'= 

I 

— 

ms 

1 

— 

m 

1 

— 

m 

VI 

r 

ns 

r 

— 

ns 

»  •  • 


•  •  • 


1  -n 


1  -n 


1  - 


n 


(11) 
(12) 

(13). 


Had  we  represented  the  polynomial  a;  +  y  -f  ^  hy  5,  the  alge- 
braic work  would  be  slightly  diminished,  but  the  preceding  method 
has  been  followed  in  order  to  show  more  clearly  the  process  of  solu- 
tion. 

30.  Find  the  values  of  the  estates  of  six  persons,  A,  B,  C,  D,  E, 
F,  from  the  following  conditions  :  1st.  The  sum  of  the  estates  of  A 
and  B  is  equal  to  a  ;  that  of  C  and  D  is  equal  to  b ;  and  that  of  E 
and  F  is  equal  to  c.  2d.  The  estate  of  A  is  worth  m  times  that  of 
C ;  the  estate  of  D  is  worth  n  times  that  of  E,  and  the  estate  of  F 
is  worth  p  times  that  of  B. 

\st  Solution. 
Le*     X     denote  the  value  of  A's  estate  ; 
a^x  '' 

y 

b-y  " 
z 
c-z  '' 


B's 

(( 

C's 

(( 

D's 

It. 

E's 

u 

F's 

{( 

from  the  conditions  of  the  problem, 


115] 


EQUATIONS    OF    THE    FIRST    DEGREE. 


117 


'  X  —  ray  =  0  •  • 
y  ■\-  nz  ■=.  h  '  • 
px  —  z  :=  ap  —  c 


»  —  my 

h  —  y  =.  nz  \     OY^ 

c  —  z  =  p{a  —  x)\ 

combining  (1)  and   (2), 

X  +  mnz  =  hm     ....     (4) ; 
combining  (3)  and   (4),  and  finding  the  value  of  z, 

bmp  -\-  c  —  ap       ,  .  ,    ,  ,       „ 

z  = ; — — -^,  which  denote  by  P ; 

mnp  +1 

combining  (3)  and  (4),  and  finding  the  value  of  x, 

amnp  -\-  bm  —  cmn       ,  .  ,    ,  ,       „ 

-,  which  denote  by  Q\ 


X  = 


mnp  -\-  1 

substituting  the  last  value  in  (1),  and  finding  the  value  of  y, 
anp  -\-  b  —  en 


y  = 


whence. 


mnp  +  1 

V»11H^11    Kl 

A's  estate 

is  equal  to 

B's     " 

"         "     c 

C's     " 

D's    " 

E's    " 

Fs    «' 

( 

which  denote  by  E ; 


(1) 
(2) 
(8) 


"         "         "     a~Q, 
"        «        "  R, 

"     b-B, 

"        "        "  ^, 

"     c  -  P. 

We  might  have  combined  (2)  and  (3),  eliminating  z,  and  then  from 
this  resulting  equation,  taken  with  (1),  have  found  the  value  of* 
ndy. 

2c?.  Solution 

By  a  single  unknown  quantity, 

Let  X  denote  the  value  of  C's  estate ; 

then  will  h  —  x  "  «  D's       « 


118  KEY    TO    DA  VIES'    BOURDON.  [127. 

w,x  denote  the  value  of  A's       " 

a  —  mx  "  "  B's        " 

p{a  —  vix)         "  «  F's       '■• 

c  —  p{a-mx) "  "  E's       « 

and  from  the  remaining  condition    of  the  problem, 
b  —  X  =:  n[c  —  p  {a  —  mx)]  ; 

whence,  by  the  rule  for  solving  equations  of  the  first  degree, 

b  -f  anp  —  nc 

X  =z  • 

rnnp  +  1 

Having  found  the  value  of  C's  estate,  the  remaining  quantities 
may  be  found  by  substituting  it  in  the  expressions  of  the  data,  and 
reducing.     The  operations  are  obvious. 

INEQUALITIES. 

1.  Given,         5a;  —  6  >  19,         to  find  the  smallest  limit  of  x. 
If  we  add  6  to  both  numbers  of  the  inequality,  we  have 

5ar  >  19  +  6,     or,     5z  >  25  ; 
dividing  both  numbers  by  5,  ■*Fe  have 

a;  >5. 

14 

2.  Given,         3z  +  —  a;  —  30  >  10,     to  find  the  least  limit  of  x. 

Reducing,  3a;  +  7j:- 30  >  10,     or,     10a:  —  30  >  10  ; 

adding  30  to  both  members  of  the  inequality,  and  dividing  by  10, 
we  have, 

a?>4. 

X    X    X     IS     n 

3.  Giver.  u— o+o  ^"77^  IT'  ^^  ^"^  *^^  ^^^^^  ™it  of  x. 

V        o         /i  -i  Z 


127-132.]  SQIJAKE    ROOT    OF    NUMBERS.  119 

Multiplying  both  members  by  6,  we  have, 

2  — 2a;  +  3a;  +  39  >  51; 

reducing,  subtracting  39  from  both  members,  and  dividing  by  5J, 

we  have, 

a;  >  6. 

4.  Given,         -—  +  ia;  —  a6  >  -—  to  find  the  least  limit  of  x, 

5  5 

Multiplying  both  members  by  5,  we  have 

ax  +  56a;  —  bab  >  a^ ; 

adding  +  5a6  to  both  members,  and  dividing  by  the  co- efficients  ot 
T,  we  have, 

(a  -f-  56)  x^  a  {a  +  56) ;     or,     x  >  a. 

bx  62 

5.  Given,         — aar  -f-  a6  <  — ,  to  find  the  largest  limit  of  x. 

Multiplying  both  members  by  7,  adding  —  7a6,  and  dividing  by 
the  co-efficients  of  ar,  w.e  have, 

(6  —  7a)  a;  <  6  (6  —  7a)     or,     a;  <  6. 

SQUARE    ROOT    OF    NUMBERS. 
Art.  104,  p.  132. 


(1.) 

72^25 

85. 

Ans. 

(2.) 
1'76'89    133. 

Ans. 

64 

)  8  25 

1 

23)  76 

8  25 

69 

0 

263  )  7  89 

7  89 

120  KEY    TO    DA  VIES'    BOUEDON.  [132. 


Ans. 


(3.) 
99^40^09  997.  Ans. 

(4.) 

85^67^8973 

9256  +  . 

81 

81 

189  )  18  40 

182  )  4  67 

17  01 

3  64 

1987  )  1  39  09 

1845  )  1  03  89 

139  09 

92  25 

0 

18406  )  11  64  73 
11  04  36 

60  37. 

Eem. 

(5.) 

67^8r2675 

(6.) 

8234  +  .  Ans.         279^24^01  1671  + 

64 
162  )  3  81 

1 
26  )  1  79 

3  24 

156 

1643  )  57  26 

327  )  23  24 

49  29 

22  89 

16464  )  7  97  75 

3341  )  35  01 

6  58  56 

33  41 

1  39  19. 

Eem. 

160. 

Eem. 

37^49^60^42 

(7.) 
6123 +  .  Ans. 

36 

121  ) 

149 
121 

1222 

)  28  60 

24:  U 

12243  )  4  16  42 

3  67  29 

1 

Ans. 


49  13.    Eem. 


132.]  SQUAKE    ROOT    OF    NCMBEES.  121 

(8.) 

36'61'09'70'49  |  60507.  Aiis. 

36 

1205  )  61  09 
60  25 


121007  )  84  70  49 
84  70  49 

(9.) 
91^87'41'67'27'04  |  958510 +  .  Ans, 

81 


185  )  10  87 

9  25 

1908  )  1  62  41 

152  64 

19165  )  9  77  67 

9  58  25 

191701)  19  42  27 

1917  01 

1917020  )  25  26  04.  Eem. 

(10.) 
74'68'41'64  |  8642.  Ans. 
64_ 

166  )  10  68 
9  96 


1724  )  72  41 

68  96 


17282  )  3  45  64 
3  45  64 


13^  KEY    TO    DAYIES'    BOURDOX.  [133-135. 


Art. 

lOa,  p.  133. 

40^96 

64 

58^2r69 

763 

36 

49 

124  )  4  96 

146  )  9  21 

4  96 

8  76 

0 

1523  )  45  69 

45  69 

•••        ^^ 

Ans. 

(2.) 

27^35^29 

523 

95^64^84 

978 

25 

81 

102  )  2  35 

187  )  14  64 

2  04 

13  09 

1049  )  31  29 

1948  )  1  55  84 

3129 

155  84 
(3.) 

•••  m- 

Ans. 

13^69    37 

a/25  = 

-.  5 

c 

1 

f 

67  )  4  69 

4  69 

/.    ^  =  7f.    Ans. 

ArU 

107,  p.  135. 

(2.) 

,'- 

11  X  225  =  2475    49 

• 

16 
89  )  8  75 

. 

8  01         /.  tl 

=  33^. 

Ans. 

74.    Rem. 

136.]  SQUARE    ROOT    OF    NUMBERS.  123 


(3.) 

223  X  1600  =  35^68^00  597 

25 

109  )  10  68 

9  81 

1187  )  87  00 

83  09  .-.  W  =  l^J 

3  91.  Rem. 

Art.  107*,  p.  136. 

(2.) 

29.^00^00  5.38.  Ans. 

25 

103  )  4  00 

3  09 

1068  )  91  00 

80  48 

10  52.  Rem. 

(3.) 

2^27.^00^00^00^00  15.0665.  Ans. 

1 

25  )  1  27 

125 

3006  )  2  00  00 

1  80  36 

30126  )  19  64  00 

18  07  56 

301325  )  1  56  44  00 

1  50  66  25 

5  77  75.  Rem. 

Ans. 


124  KEY    TO    DAVIES'    BOUEDON.  [136-137. 


(4.) 

3;00^00'00  1.732.  Ans. 

1 

27  )  2  00 

1  89 

343  )  11  00 

10  29 

3462  )  71  00 

69  24 

1  76.  Eem. 

Art.  108*,  p.  136. 

(1.) 

3271.^47^07  57.19.  Ans. 

25 

107  )  7  71 

7  49 

1141  )  22  47 

11  41 

11429  )  11  06  07 

10  28  61 

77  46.  Mem, 

(2.) 

31.^0270  5.57.  Ans, 

25 

105  )  6  02 

5  25 

1107  )  77  70 

77  49 

21.  Rem. 

137-140.]  SQUARE    ROOT    OF    POLYNOMIALS.  135 

(3.) 
O.OI'OOIO^OO'OO  I  .10004.     Ans. 
1 

20004  )  0  00  10  00  00 
8  0016 

1  99  84.     Re77i. 


SQUARE    ROOT    OF    POLYNOMIALS. 

25a*—d0a^  +  4:Qa'b^—24:ab^—lGb^  \5^—Sab+^.  A7is. 

25a* 

lOa^—Sab)  —■S0a^  +  4:9a^^ 


10fl3— 6«i4-4J2 )  40a2Z)2_24a^>3  +  16^)* 
40a2^,2_24g63-^16&* 

0 

Note. — The  practical  operations  are  carried  on  in  the  same  way  as  in 
extracting  the  square  root  of  numbers.  Having  found  5a^  we  double  it 
and  divide  — 30a^6  by  the  result,  writing  the  quotient  both  in  the  divisor 
and  in  the  root,  and  so  on  to  the  end. 

(2.) 

^4 ^ 4„3a; ^_ ^(^,jp. 4. 4^a;3 + a4  j  a^j^'iax-\-x\    Ans. 


2a^-^2ax)  4:a^x  +  Qa^x^ 
4:a^x  ^-  4:a^x^ 


2a2  +  iax  +  x^ )  'Mx^  +  ^aa^  +  «* 

0 


126  KEY    TO    DAVIES'    BOURDON.  [140. 

(3.) 


a^ 

2a^—ax)  —'Mx^ZaH^ 

—  2a%+   a^x^ 

2a^—2ax  +  x^ )  2a^x'^—2ax^  +  3^ 

2a^x^—2aa^  +  x* 

(4.) 

42;6  +  12a;5  +  5a^-2a:3_^  7a;2_3a;  +  l 

2x^-]-d3^—x  +  l.    Ans. 

4:afi 

^3^  +  dx^)12a^-\-6x* 

12a.-5  +  9a:^ 

4a^^Gx^—x)  —4:X*—2x^-\-7x^ 

_4a4_6a^+  x^ 

4a;6  +  62;2— 2a;  +  l  )  4:X^+6x^—2x  +  l 

4as^6x^—2x  +  l 

(5.) 
9a*-12a^  +  2Sa^^—16ab^+16¥  \  Sa^—2ab  +  U^  Ans. 


9a* 


ea^—2ab)  —l2a^b  +  2Sa^'^ 
—  12a%+  ^aW 


6a2-4a5  +  452 )  2^aW—\^ab^^\m 
24a2&2_i6g^  +  16&^ 

(6.) 

3d  divisor.                                   Boot  =  5a^b—4abc  +  Qbc^~3a''c.  Ans. 
10a?b-8aic  +  Gftc' )  QOa^b^c^-  48a6V  +  SQb'c^-SOa^bc  +  24a'&c'^  -  dQa'bc^  +  9a*c'^ 
60a''6^c''-48a6-'c»  +  366°c« 

10a^b-Sabc  +  12bc^-3a''c )  -d{)a*bc  +  24a^bc'-3Qa^b(^  +  9a*c^ 

-  30a*bc  +  24^36^2  _  ^Qa''b(^  +  Qa*c^ 


140.]  SQUARE    ROOT    OF    POLYNOMIALS.  127 

Explanation. — In  this  example  we  find  by  the  rule  that  the  first  two 
terms  of  the  root  are  5a'6— 4cfftc,  and  that  the  second  remainder  is 
SOaWc"^—,  etc.,  as  above  written.  We  then  continue  the  operation,  as 
shown  above.  This  method  of  proceeding  is  necessary  to  get  the  work  on 
the  page, 

(7.) 

x^  +  2ax^  +  3a^x^  +  2a^x  +  a^  \x^  +  ax  +  aJ^.    Am. 

«* 

2x^-hax)  2ax^  +  3alv^ 
2aot?  +    a^x^ 
2x2  ^  2ax  +  a?  )  2a^x^  +  2a^x  +  a* 
2a^x'^  4-  2a^x  +  a* 

(8.) 

25x^1/— 30a^y^  +  29xY—12xy^-\-^f  \  5x^y—3xy^-\-2f. 

25x*y^ Ans. 

10a^y—3xy^  )  — 30a;y  +  5i9a:Y 
— 30a:Y+   9zY 


10x^y—6xy^  +  2y^ )  20xiy*—12xy^  +  4:f 


(9.) 
a*  _  4a;5  _^  4^4  _^  12^:3  _  24a;2  ^  ^q  |^^3__2a^^,  ^^^^ 


2x^  —  2x^)  —Ax^-\-  4a^ 

2a;3  —  4^2  +  6  )  122;S  —  24^2  +  36 


123;3  _  24.T2  -^  36 


128  KEY    TO    DAVIES'    BOUEDON.  [141. 

(10.) 

xi  ^2a^  +  ^x^  —  \x  +  -^  I  x^  —  0-  +  h    ^ns. 

3^ 


2x^  —  x)  —  2x^  +  |a;2 
—  2x^+    x^ 


'Zx^-'Zx  +  l  )  ia;2_i2;  +  ^ 
^x^  —  ^a;  +  -jV 


(11.) 

a;2  +  fax  +  ^a^  —  Z>a;  —  ^a§  +  ^^J^  |  a:  +  ^a  —  ^5.     Jws. 
a;2 


^ax  +  ia2 


2a;  +  fa  —  |J  )  —  to  —  ^a^'  +  W 
—  hx  —  ^ah  +  i?^ 


(12.) 

4:a:*  +  6a:3  +  ^x^  +  15a:  +  25  [  ^x^  +  l-^  +  5.     Ans. 
^x* 


4a;2  +  fa; )  6x^  +  -V^'^ 
6a,-3  +  f  a;2 


42;2  +  3a;  +  5  )  20a;2  +  15a;  +  25 
20a;2  +  15a;  +  25 


(13.) 


^2m 


2am  _  253n  )   _  4a°»J3n  _|.  4J6Q 
_  4^<mj3n  +  456n 


141-145.]      EADICALS    OF    THE    SECOND    DEGEEE.  I5i9 

(14.) 


i«* 

c? 

+  ax 

)  a^x 

+ 

fa2^'2 

a?x 

+ 

a2a;2 

a2  4-  %ax  +  ^.r2 )  1^23,2  ^  2Qja;3  +  ^^4 


(15.) 

«2  +  2a;  —  1  —  2a;-i  +  a;-^  I  a;  +  1  —  a:-i  =  a;  +  1 ^W5. 

'                                                                                                     X 
^  I 


2a;  +  1 )  2a;  —  1 

%x  +  1 


2a; +  2 -a;-!)—  2  —  2a;-i  +  a;-2 
—  2  —  2a;-i  +  a;-2 


RADICALS    OF    THE    SECOND    DEGREE. 
Art,  113,  p.  145. 

2.  Vi2855^  =  ^mkKfyJb  =  Sb^aMV^.    Ans. 

3.  V^^^Wc  =  VlQa^^  X  2ac  =  4a*^V2ac.    Jws. 

4.  V'2o6a2J^c8  =  16ffJ2^.     J«5. 

5.  Vl024a9^;V5-  _  Vl024a8*V  x  «Jc  =  d2a^b^cWabc.    Ans. 

6.  Vn^a>¥^  =  ^/lamfyCmabd 

=  2aW(^VlS2aM.    Ans. 


130  KEY    TO    DAVIES'    BOUKDOK.  [145-148. 

7.     y/ab  (a^  +  ^ab  +  4^  =  («  +  2i)  Vab.    Am. 


8.     V2  (aa;  —  W)  {a^—hx^f  —  (a—bx^)  \/2  {ax—bx^).    Ans. 


9.     V{x  —  4)  (a;2  +  4a;  +  4)  =  (a;  +  2)  Va;  —  4.    Ans. 

Art.  115,  p.  147. 

2.  7V2a  +  3  V2a  +  2V2a  =  12\/2a.    Ans. 

3.  Vie^  X  3a  +  ^iA/25x3fl  =  (4^>  +  5^>)  A/3a 

=  9bV^a.     Ans. 

4.  7\/2a  —  2V2a  =  (T  —  2)  \/2a  =  5V2a.     Ans. 

5.  \/l652  X  3a  —  bV25  x  3a  =  (ib  —  5b)  VSa 

=  —  b\/3a.    Ans. 

6.  V49a*62  X  2a  —  VOft^Fx^  =  {7a^  —  3a¥)V2a 

=  (7a  —  3b)abV2a.    Ans. 

Art.  116,  pp.  147-148. 


3.  2a  X  -  3a\/(a2  +  b^^  =  —  Ga^  (a^  +  l^).    Ans. 

4.  ^—'Z^'^—3\/^-^eV^==a  —  5Vab+6b.Ans. 

5.  15\/40  +  6\/25  — lOVie  — 4\/i0 

=  15  X  2\/l0  +  6x5  —  10x4-  WlO 

=  _  40  +  (30  —  4)a/10  =  —  40  +  26\/T0.  Ans. 


6.     2a  X  3J  X  aVic  x  abc  x2a  —  Qa^bVb^c^a^  x  2 


148-150.]  RATIONAL    DENOMINATORS.  131 

Art.  117,  p.  14:8. 

-      12ac      /Qbc       o     /5~      ^ 
4c    V    'Zb 

2ba^b      Imhi  ha      Irr?  5am      /I 

hab'^    V   mn^        b    \J   n   '~     b     \J  c 

V  (a2  —  2ac  +  c2)  X  5c       a—  c\i  h 


1.     - 


RATIONAL    DENOMINATORS. 
Art,  118,  p.  149. 

7  X  (3  +  \/5)  21  +  7v'5        21  +  7\/5 


2. 


(3  _  V5)  (3  +  Vo)  9-5  4 

=  9  to  within  :|^.    Ans. 

7a/5  X  (VTT— V3)  7a/55  — 7a/15 


3. 


(vH  +  V3)  (Vii  -  V3)  1^  -  3 

_  7\/55  —  7\/l5 
~  8 

=  31  to  within  -g^.     ^w«. 

(3  +  2^7)  (5\/l2  +  6\/5) 
(5\/l2  —  6\/5)  (5a/12  +  6\/5) 


132  KEY    TO    DAVIES'    BOURDON.  [150. 

4.  (3  +  V3)(3  +  V5Hv;5-2)^  ^^^^  ^^^^ 

(5-V5)(l  +  V3) 
5  +  V 5     and     1  -  v^  5    ^l^is  gives, 

(3  +  V3)  (1  -  V3)  (3  +  V5)  (5  +  a/5)  (V5  -  2). 
(5  -  a/5)  (5  +  V5)  (1  +  a/3)  (1  -  a/3) 

which,  after  performing  the  operations  indicated,  and  reducing, 
becomes, 

—  40  a/15  —  80  a/3  +  80  a/3  +  32  a/15       1   /— 

m^2 =  r^^  ■ 

_     Va  +  X  -\-  a/a  —  X      Tvf  ,, .  1  .      ,    ,,    ,  ,         / 

5. Multiplymg  both  terms  by   \a  +  z 

ya  -\-  X  —  w  a  —  x 


4-  -A/a  —  X,  we  have, 


2a  +  2  A/a^  —  x^  _  a  +  a/^^  —  x^  ^ 
%x  ~  X 

(7  _2  a/5)  +  2  +  a/45        ,;r  1^-  1  •         u  XI      X 
6.  ^^ /     — ^-^^ — .       Multiplying    both    terms    by 

2  a/3 -a/7  -^ 

2  a/3  +  a/7,  the  denominator  becomes  5.    For  the  numerator, 
performing  the  multiplications  and  reductions,  we  have, 

7  —  2  a/5"+  2  +  a/45 
2a/3+  a/7 


14  a/3  —  4  a/15  +  4  a/3  +  2  a/135 

+  7  a/7  —  2  a/35  +  2  a/7  +  a/315. 

But,       2  a/135  =  6  a/15  ;    and     a/315  =  3  a/35; 
hence,  the  product  reduces  to 

18  v/3  +  2  a/15  +  a/35  +  9  a/7. 


150-151.]  MISCELLANEOUS    EXAMPLES.  133 

But, 

18^/3  =  9  X  2a/3;    and    2  a/15  =  2  \/3  x  VH ; 
also, 

a/35  =  a/7  X  a/s; 
hence,  the  sum  reduces  to 

ft  

9x2'v/3  +  2\/3xa/5  +  a/7x  a/5  +  9\/7 

=  9(2a/3  +  V7)-f  VS  (2^/3  + a/7) 
=  (9  + V^)(2v^  +  a/7). 

MISCELLANEOUS    EXAMPLES. 
1.     a/125  =  a/25  X  5  =  5  a/5,     ^ws. 

3.  A/98fl2^  =  A/49a2  x  ^  =  laV^x.    Ans. 

4.  y/x^  —  (i^z^  =  A^a;2  x{x  —  a^)  =  x'^x  —  (]?■.     Ans. 

5.  a/36  X  2  +  a/64  x  2  =  Ga/2  +  8a/2  =  14a/2.    Jws. 

6.  A/9"^r3  +  a/49  X  3  =  3^3  +  7 a/3  =  10 a/3,    ^ws. 


7.  A/i  X  6  +  a/tIo-  X  6  =  iA/6  +  t^a/6  =  HV6.  ^*w. 

8.  2Vcfx~^  +  3A/64a;*  x  T  ==  2aA/^  +  24a:V^' 

=  (2a  +  24a:2)^^.    j^^s. 

9.  9a/81  X  3  +  10a/121  x  3  =  81a/3  +  110a/3 

=  191  a/3.    Ans. 

10.  a/^  X  15 -a/^  X  15  =  ^V15_^a/15 

=  ;j^a/15.     Ans. 

11.  5  X  3a/8  X  5  =  15 a/4  x  10  =  30a/10.    ^M5. 


13.     I  X  |A/i  X  ^V  =  iVS^  =  iViio  X  35 


134  KEY    TO    DAVIES'    BOURDON.  [151-155. 

13.     fV^  =  SV^.     Ans. 


14.     VQa*b^  X  2ab+V2ba^/^  x  2ab  =  Za^'s/'Zab  +  hab^%ab 

=  (3fl2i  +  5a J)  V2a^.    J?is. 


15.     4A/r><~3l  -  5A/i  X  2  +  12Vi  X  3 

=  f  a/21  -  |a/2  +  -VV3 

=  |V2I-Jg^V3  +  ¥^/3 

=  i(8V3l-15\/2  +  36V3).    Ans, 

INCOMPLETE    EQUATIONS    OF   THE    SECOND    DEGREE. 
Art,  122,  pp.  154^155. 

2.  Givea    ^x^  =  5,    to  find  x'  and  x". 
Reducing,  a;^  =  |.  =  i.  x  15. 

.-.     x'  =  lVl5    and    x"  =  —  ^\/l5.     Ans. 

3.  Given     11  (a;^  _  4)  =  5  {x^  +  2),     to  find  x'  and  a;". 
Performing  multiplications,     lla;^  _  44  =  5x^  +  10; 
Transposing  and  reducing,     (Jx^  =  54,     or    x^  =  9. 

.-.    x'  =  3,     x"  =  —  3.     Jws. 

4.  Given     —  =  w,     to  find  a;'  and  x". 

X 


Clearing  of  fractions  and  squaring, 

m^  —  x^  = 
Transposing  and  reducing, 


m^  —  x^  =  mlr^ ; 


x^  = 


m^ 


1  4-^2 
^         x"  =  _ 


X  =  — ,        x"  = : Ans. 


155.]  INCOMPLETE    EQUATIONS.  135 

5.  Given    9.^2  _^  9  =  2,x^  +  63,    to  find  x'  and  x". 
Transposing  and  reducing, 

6a;2  =  54, 
or  a;2  _    9  J  ^ 

.-.    x  =  3,        x"  =  —  3.    Ans. 

6.  Given    8  \x  +  -)  =  —  —  3,     to  find  x  and  x". 

Reducing,        56.^2  -f  56  =:  21x  +  70a;2  _  21a:, 
or,  14a;2  =  56, 

or,  .-r^  =    4. 

.-.     a;'  =  3,         a:"  =  —  2.     ^»5. 

3  3 

7.  Given 1-  :; =  8,     to  find  x'  and  x". 

1  -\-  x       1  —x 

Clearing  of  fractions  and  reducing, 

3  _  3a;  +  3  +  3^;  =  8  —  8a;2,     or    8a;2  =  2,    or    x^  =  ^\ 

8.  Given    — — -„  -\ , — 5  =  — ,  to  find  x  and  x  . 

X  —  a^       X  +  a^        6 

Clearing  of  fractions, 

3  (a;2  +  ^a^x  +  «*)  _^  3  (a;2  _  ^a^x  +  a^)  —  10  (x^  —  a*). 
Reducing  to  form  x^  =  q,   —  ^x^  =  —  16aS  or  x^  =  Aa*; 

.:    x'  =  2a^,    x"  =  —  2«2.     Ans. 

9.  Given     '—^ t  =  2,  to  find  x'  and  a;". 

X  X  —  0 

Clearing  of  fractions  and  reducing, 

a^  —  2ix  +  ^  —  dx^  =  2x^  —  2bx,    or    x^  =  ^i^; 

.-.     a;'  =  ^b,    x"  =  —  ^b.     Ans. 


136  KEY    TO    DAVIES'    BOURDON.  [155-158. 

10.     Given    -z  +  " =  - ,    to  find  x'  and  x". 

ao  ex  c 

Clearing  of  fractions, 

cx^  +  «5^^  —  tt^c  =  aix^, 

or  x^  =  c^h. 

.:    x'  =  aVb,    x"  =  —  a'y/b.     Ans. 


COMPLETE    EQUATIONS    OF    THE    SECOND    DEGREE. 

Art.  123,  2n>'  158-100. 

x^       a  h  2x^ 

7.  Given         -^ jx  =  \ x — ,     to  find  the  values  jf  *. 

o        0  a  6 

Clearing  of  fractions,  transposing  and  reducing, 

o2  _  62 

x^ -. — x  =  1  ; 

ab 


whence,  by  the  rule 


a2  _  62  r        a*  —  2aW  +  6*        a^  —  b^       a^  +  b* 

X  — ■+-  \  / 1  A ■ —  =. ±  — ■ — ; 

2ab      ~V  4a262  2ub  2ab 

taking  the  upper  sign,  x  =  —r  =  -, 

262  5 

"         lower     "        X  =  —  -—7  = 

2a6  a 

c?a:       3x2  1  +  c       x^       x 

«  Given h  -r  +  1  = T  "^  y 

c  4  c  4       a 

to  find  the  values  of  x. 

Clearing  of  fractions,  transposing  and  reducing, 

rf2  _  c         1 


159.]  EQUATIONS  OF  THE    SECOND    DEGREE.  137 

whence,  bv  the  rule, 


(P  —  c  /I       d'^  -  2cd'^  +  c^  d^—  c       d^  +  c 


v/4 


"^  ~         2cd      ^  V   c  "^  4c2(/3  ~  2cd     "^    2cd     ' 

,.        u  •  2c        1 

tiiking  the  upper  sign,  x  —  ——  =  -3, 

2(^2  d 

"         lower     "        X  =z  —  -— ,  =  —  -. 

2ca  c 

9.   Given —  +  —  =  8  — -,    to  find  the  values  of  t. 

4         3         8  4        o 

Clearing  of  fractions,  transposing  and  reducing, 

2       2         5 

x^ X  =1  — i 

3         4 

whence,  by  the  rule, 


1  /5  ,    1       17 

^=3^V4  +  9  =  3^6' 


3  ^  5 

hence,  ^  =  -      and      2;  =  —  -• 

2  o 


90         90  27  ^   ^    ,      ■  ,         ^ 

10.  Given -—7  =  — — n'    to  find  the  "alues  of  x. 

X        X  +  I       X  -\-  2 

Dividing  both  members  by  9, 

10         10  3 


X        X  +  i        a:  H-  2' 
clearing  of  fractions,  transposing  and  reducing. 


2       ■''        20 

x^ X  =  — : 

3         3  ' 


whence,  by  the  rule, 


138  KEY  TO  DA  vies'  boukdon.  [169. 


7  /20       49       7       17 

X    -     -fc       4     / I _    -I-    • 

~6        V336~6~6' 

hence,  x  =  4.     and    x  =  — —  =  —  -    . 

o  o 

11.  Given         —- 2  = — ,  to  find  the  valueT  of  x. 

8  —  z  X  —  2 


Clearing  of  fractions,  &c., 


,       39  28 

x^ X  z= : 

5  5  ' 


whence  by  the  rule, 


=%-V^ 


28       1521       39       31 


+ 


5         100        10       10 


> 


hence,  x  =  7,     and     a:  =  — -  =  -. 

10       5 

a;2  b  —  I  b 

13.  Given         ax r  +  ^  = — ; —  ^"^  +  -^,  to  find  the  values  of  ar. 

0  0  a 


Clearing  of  fractions,  d^c. 


X^ X  =  ot 

a 


whence,  by  the  rule, 


a^  —  b  /        a*  —  %i^b  +  62       a?  —  b       a^  +  b 


±v/*  + 


^  =  -2r-=^V'  + 4^5 =  -2—^      2a     ' 

hence,  x  z=  a,  x= • 


13.  Given         _-.  +  ^  - -,  = -^:r  +  ^  "  ^' 
to  find  the  values  of  x. 


159.]  EQUATIONS  OF  THE  SECOND  DEGREE.  139 

Cloaring  effractions,  &c., 


X^ X  =. 


2b         a?-  —  62 

X'  — 

whence,  by  the  rule. 


c  c2       ' 


h  /^2^ 


2  _  62  ^2  5  a 


,                                      6  +  a  ,  6  —  a 

hence,  a;  = ,     and     x 


c  c 


14.  Given         mx^  -{•  mn  =  2my«  •  x  +  «.c^  to  find  the  values  of  x. 
Transposing  and  reducing, 


2m  -i/n  mn 

x^ -—  X 


m  —  n  in  —  n 

whence,  by  the  rule, 


niy/n  f         fnn     ~~       Tn^ri  m\/n         nym 

~  m  —  n       V       m  —  n       (m  —  ny       m  —  n~  m  —  n  ' 


m-\/n  +  n^/m  V"'"  VV  "* -h  v  «)  ^/mn 

nence,  x  = 

m  —  n 


since,         m  —n  —  {y/ni  +  -y/n)  (-y/m  —  ^/ny         (Art.  47.) 
also, 

y«~—  n \^in  y/mn  ( y/m—  ^/n)  __        ^/mn 


X  =  — 


2c?  Solution. 

mx^  +  mn  =  2ot  yV-  x  +  «a;2 ; 
transposing  the  first  term  of  the  second  member,  we  have 


140  KEY  TO  DAVIES'  BOUEDON.  [159. 

mx^  —  2m  y^  •  X  +  mn  =  nx^ ; 

observing  that  the  first  member  is  the  square  of  a  binomial  whose 

terms  are     ^/m  •  x  —  -y^'la  ■y/n',     we  have, 

•  ym  •  X  —  Y  «i  -y/ti  =  ±  -y/n  .  ar; 

yw  •  X  +  -yAi  •  a;  =  ^^m  -y/TT 
(ym  q=  yn)  x  =  ■y/ni'^n  :  hence, 


a;  =  — ;= and,     x  =  — :^- 


■y/m  —  -yjn  -y/m  +  y^ 

1  o.  triven         aox^  —  ■ — ■  -\ = x, 

c^         c  <?  c 

to  find  the  values  of  x. 

Clearing  of  fractions,  &c., 

,    ,   b^  +  Sa^         aft  — 252  +  6a2 
abc  abc^ 

«vhence,  by  the  rule, 


^  _  _  ^H^^  ^      /  ab  -  262  ^  6„2       6*  4-  6^252  4.  9^4 
2abc      ~  V  a6c2  '  io^feV 

__  _  62  -f-  3a2       3„2  _^  4q5  _  52 
~  2a6c      ""  2a6c  ' 

hence,   x  =  ^-^^^^!^  =  "JL=1,    ,  ..       6^^  +  4a6  ^       3a  +  2& 
2a6c  oc  2a6c  6c 

„  ^.  4a;2    ,   2a;  3^2       58ar 

16.  Given         _ -|- _  +  10  =  19  -  —  +  — , 

V)  find  the  values  of  z. 


159.]  EQUATIONS  OF  THE  SECOND  DEGREE.  141 

Clearing  of  fractions,  «Sjc., 

a;2  _  8x  =  9  ; 
whence,  by  the  rule, 

ar  =  4±^9+lG  =  4±5, 

hence,  a;  zn  9,     x  —  —  1, 

17.  Given         i  = 1    to  find  the  values  of  x. 

X  —  a  a  +  X 

Clearing  of  fractions,  &c., 

-     6-2     ' 
whence,  by  extracting  the  square  root  of  both  members. 


/b  +  2  ,  /T+ 


18.   Given         2a;  +  2  =  24  —  5x  —  2*2^  to  find  the  values  of  x 
Transposing  and  reducing, 


a-2  +  ^x=ll; 


whence,  by  the  rule, 


X 


7  /,    ,49  7       15 


whence,  ;p  =  2,     and     x  = ^• 

19.  Given         x'  —  a:  —  40  =  170,     to  find  the  values  of  x. 
Transposing, 

a;2  _  a;  =  210 } 


143  KEY  TO  DAVIES'  BOURDON.  [159. 

whence,  by  the  rule, 


.  =  1.^210  +  1=  i±|; 

hence,  x  =  15,     and     a:  =  —  14. 

20.  Given         Sx^  +  2z  —  9  =  76,     to  find  the  values  of  «. 
Transposing  and  reducing, 


2,2         85 


whence,  by  the  rule, 


X 


1  /85   .    1  1        10 

-4-    *  / — -fc  —  • 

3~V3^9  3       3' 


17  2 

hence,  x  =  5,     and     z  = -=  —  5  -• 

o  o 


21.  Given         a^  -\-  b^  —  2bx  -\-  x^  ■=.  — —,  to  find  the  values  ol  a? 


w2 


Clearing  of  fractions,  &;c., 


25n2  rfia^  +  nW 

«^  +  —. -.X  = 


m^  —  n^  rn?  —  n"^    ^ 


whence,  by  the  rule. 


*  = ±  \/ ?■— -^-  +  —^ 


7^2  — ^2        Y     ;^2  _  ^2  ^4  —  2m^n^  +  n* 


bn^            n  %/  a?ra'^  -\-  h'^m?  —  d^v?- 
— ^  — ~ • 

whence,         x  =  ^^  _  ^^   |  bn  ±  ^a^m^  +  62^2  —  aV-"  [ 


160.]  EQUATIONS    OF    THE    SECOND    DEGREE.  143 

1  3  1 

22.  Given  ^ +  -, =  - ; 

2{x — 1)       x^  —  1       4 

whence,  2  (a:  +  1)  +  12  =  a:^  —  1, 

and  . •.     x  =  l  ±  Vis,    x'  =  5,    x"  =  —  3. 

23.  Given         -  +  --^^^-^  =  ^-^ ; 

whence,     19.?;  (10  —  x)  +  3800  =  9  (10  +  x)  (10  —  x). 
Reducing,  a;2  —  19a;  =  290 ; 

1  19   ,    ./C7T"361       19  ±39 

hence,  a;  =  y±y290  +  —  =  — ^ —   • 

.-.    a;'=29,     a;"=  —  10. 

/j;2 Q^  1 

24.  Given =  a;  —  3  +  - ; 

a;  +  3  ^  x' 

whence,  x{x^  —  6x)  =  (x^  —  9)a;  +  a;  +  3. 

8  3 

5  5' 


Reducing,  x^  —  k  ^  =  — 


4±1         ,      ,        „      3 
or  a;  =  — =— ;     a;  =  l,    a;  =-; 

5  0 

25.  Given =  x  H ; 

a;  —  1       2  a; 

whence,  2a;2  =  3  (a;2  —  a;)  +  2  (a;2  —  2a;  +  1) ; 


hence,    a;2_-a;=-3,        or        x  =  -±V^q-:^; 

1 

.'.    a;' =  2        and        x"  = -. 

o 

na    r-                      a;  +  2a;  — 2       5 
26.  Given  ^r — -  =  - ; 

a;  — 2       a;  +  2       6' 

whence,     &  {a?  +  ^x  +  A)  —  %  [x^  —  4.x  +  ^)  =  h  {a?  —  ^)) 

48 
therefore,  x^ ^  a;  =  4 ; 


144  KEY   TO   DAVIES'    B0UKD0N.*-'^1{'-3f  1,16Q 


AX.    ,u        1  24^./5767~     24±36 

and  by  the  rule,    z=—±y  -^  +  4= — - —  ; 

2 

.*.     a;'  =  10        and        x"  =  —  -. 

5 

o^    n^  a: -6       a; -12       5 

27.  Given ^=«; 

a;  —  12       X  —  6       6 

■whence,      6(a;2— 12a;  +  36)  — 6(a;2-24a;  +  144)=5(a;2— 18a;  +  72) 

Reducmg,  x^ =-  x= — ; 

o  o 


81        ,  /6561       5040       81  ±  39 
and  ^^_±y____-  =  __^_: 

42 
.•.    a;' =  24        and        x"  =  -^. 

0 

^r.    r^-  X  a;  +  1       13 

28.  Given  — —^  -\ -^—  =  -^ ; 

X  +  1  X  6 

whence,          6a:2  _^  6  (a;2  +  2a;  +  1)  =  13  {x^  +  x) ; 
hence,  a;  =  —  ^  ±  V^  = ^— ; 


29.  Given 


a;'=2        and        a;"=:— 3. 
1  2  3 


a;  —  2       a;  +  2       5 ' 
whence,  5  (a;  +  2)  -  10  (a;  -  2)  =  3  (a;^  -  4) ; 

5       .  /7.       25            5  ±  23 
hence,  x=  —-±y  14  +  ^= ^ — ; 

14: 
.*.    a:  =  3        and        x= ^. 

4  5  12 

30.  Given  — — r  + 


a;  +  l^a:-i-2~"a;  +  3' 
whence,  4(a;2  +  5a;  +  6)  +  5(a;2  +  4a;  +  3)=12(a;2  +  3a:  +  2); 

2       ,/4      45       2±7 
hence,  a;  =  g±|^^  +  y=     3     5 

5 

.♦.    x'=  3        and        a;"=  —  5. 


162.]  EQUATIONS   OF  THE   SECOND   DEGREE,  145 

PROBLEMS  GIVING  RISE  TO  EQUATIONS  OF  THE  SECOND  DEGREE 

4.  A  grazier  bought  as  many  sheep  as  cost  hitn  £60,  and  after 
reserving  15  out  of  the  number,  he  sold  the  remainder  for  £54,  and 
gained  2s.  a  head  on  those  he  sold  :  how  many  did  he  buy  1 

Let    z     denote  the  number  purchased  : 
and  z  —  15,     the  number  sold  ; 

then  will       denote  the  number  of  shillings  paid  for  1  sheep, 

and  the  number  of  shillings  received  for  each. 

X  —  15 


From  the  conditions  of  the  problem, 

1200  _    1080 
z     ~  X  —  15 

clearing  of  fractions,  &;c., 

z^  +  45a;  z=  9000 

whence,  by  the  rule. 


2; 


45  /  '       '     ' 

z  =  -  —  ±  W  9000  + 


2025  _       45       195 


4  2  2' 

hence.  x  =  15,     and     a;  =  —  120, 

the  positive  value  only,  corresponds  to  the  required  solution. 

5.  A  merchant  bought  cloth  for  which  he  paid  £33  15s.,  which  he 
sold  again  at  £2  8s,  per  piece,  and  gained  by  the  bargain  as  much 
as  one  piece  cost  him  :  how  many  pieces  did  he  buy  1 

Let     X     denote  the  number  of  pieces  purchased  ; 

675 

then  will,         ■     denote   he  number  of  shillings  paid  for  each. 


146  KEY  TO  DAVIKS'  BODKDON.  [163. 

ftnd  48.J:  the  number  of  shillings  for  which  he  sold  trie  whole. 
From  the  conditions  of  the  problem, 


48. -,675=^ 


X    ' 


then,  by  clearing  of  fractions,  &c., 

225         225 


X^ tttX  — 


16  16 

whence,  by  the  rule. 


_  225  /  225       506i 

^~"32  ^  V  ~16""^T02 


50625  __  225       255 
024~  ~  32  ^"32'' 


using  the  positive  value  only,         x  =  — —  i=  15. 

6.  What  number  is  that,  which  being  divided  by  the  product  of 
its  digits,  the  quotient  will  be  3 ;  and  if  18  be  added  to  it,  the  ordei 
of  its  digits  will  be  reversed  ? 

Let     X     denote  the  first  digit, 
and,        y         "  second  " 

then  will  10a;  +  y         denote  the  number. 

From  the  conditions  of  the  problem, 

10-^  +  y  ^  3 

10a;  +  2/  + 18  =  lOy+ar; 

whence,  by  reduction, 

lOx -\- y  =  Sxjf, 

y  —  x  =2; 

finding  the  value  of  x   in  terms  of   y   from  the  second,  and  sub 
stituting  in  the  first,  we  have. 


163.]  EQUATIONS    OF   THE    SECOND    DEGREE.  147 

lOy  —  20  +  2/  =  3y2  -  6y  ; 
whence,  by  transposing,  &c., 


o       17  ,  20 


by  the  rule, 


17  /      20      289  17 


3    '     36         '    6   ~6' 


taking  the  positive  sign,  y  =  4  ; 

whence,  a.  ^  2,     and  the  number  is  24. 

7.  Find  a  number  such  that  if  you  subtract  it  from  10,  and  mul 
tiply  the  remainder  by  the  number  itself,  the  product  will  be  21 . 

Let     X     denote  the  number : 

from  the  conditions  of  the  problem, 

{\0  -x)  x  =  2\;     or,     x^^  -  \0x  z=  - 'Z\  ; 

by  the  rule, 

z  =  5  ±  y  -  21  +  25  =  5  ±  2; 

whence,  x.  =  7,     and     a;  =  3. 

8.  Two  persons,  A  and  B,  departed  from  different  places  at  the 
same  time,  and  travelled  towards  each  other.  On  meeting,  it  ap- 
peared that  A  had  travelled  18  miles  more  than  B;  and  that  A 
could  have  performed  B's  journey  in  15f  days,  but  B  would  hav€ 
been  28  days  in  performing  A's  journey.     How  far  did  each  travel  1 

Let       X        denote  the  number  of  miles  B  travelled ; 
X  +  18         "  "  "     A         « 

X 


M 


15| 


A         "         in  one  day  j 


148  KEY  TO  DAVIES'  BOURDON.  [163. 

a?  +  18 


28 

re  +  18 


denote  the  number  of  miles  B  travelled  in  one  day , 
"  "  days  A         « 


X 
«  «  t;    g  M 


(X  +  18\ 
\     28     I 


28 
from  the  conditions  of  the  problem, 

a;  +  18  X  a;2  +  36x  +  324        z^ 

=  r-T  -r^        or, 


X 


15f  28 


a;+  18  '  28  ~  15| ' 


clearing  of  fractions,  and  reducing, 

324         2916 


ar^ — -  X  = 


7  7 

By  the  rule, 


162  Ml6 

=  -7-=^V-7- 


162      216 


26244 


a;  =  -—  d= 


7    ~    7    * 

«ence,  using  the  upper  sign, 

378 
x  =  ——  =54. 

7 

9.  A  company  at  a  tavern  had  £8  15*.  to  pay  for  their  reckoning; 
but  before  the  bill  vpas  settled,  two  of  them  left  the  room,  and  then 
those  whc  iemained  had  10s.  apiece  more  to  pay  than  before:  how 
many  were  there  in  the  company  ? 

Let  X  denote  tne  number  in  the  company. 


163.]  EQUATIONS  OF  THE  SFCOND  DEGREE.  149 

175 

Then      —   will  denote  the  number  of  shillings  each  should  pay  j 

175 

"  '•  «  "    paid; 


X  —  2 
from  the  conditions  of  the  problem, 


175  175       ,^ 

-  —  =  10; 


X  —  2         X 
clearing  of  fractions, 

175a;  -  175a;  +  350  =  lOx^  -  20x; 
whence,  x^  —  2x  =  35. 

By  the  rule, 

X  =  1  ±^36=  1  ±  6; 


using  the  upper  sign,         x  =7. 

10.  What  two  numbers  are  those  whose  difference  is  15,  and  o» 
which  the  cube  of  the  lesser  is  equal  to  half  their  product  ? 

Let       X       denote  the  smaller  number; 
then  will  a;  +  15  "  greater         " 

from  the  conditions  of  the  problem, 

a;3  =  1  (a;2  +  15a;),     or,     x^  =  l{x  +  15) ; 
Py  the  rule, 


1  /15   .     I        1       11 

4V2^16      44' 

using  the  upper  sign,         x  =  S  ;         hence,         a;  +  15  =  18. 
11,  Two  partners,  A  and  B,  gained  $140  in  trade:  A's  money 


150  KEY  TO  DA  vies'  BOURDON.  [163. 

was  3  months  in  trade,  and  his  gain  was  $60  less  than  his  stock :  B"s 
money  was  $50  more  than  A's,  and  was  in  trade  5  months:  what 
was  A's  stock? 

^-•^^         X  denote  the  number  of  dollars  in  A's  stock  j 

a:  +  50  "  "  "  B's     " 

X  —  60  "     A's  total  gain  ; 

«  -  60  ,,      , , 

As  gam  per  month  ; 


3 

g-  60 
3a; 


"     A's     "         "        "    on  1  dollar ; 


'g         «  U  It 


U  —  60\ 

I  — g^  I  (•'^  +  50)  5  B's  total  gain 


From  the  conditions  of  the  problem, 


«- 60  +fc^^-^  {x  +  50)  5  =  140; 


clearing  of  fractions,  and  reducing. 


a:2-i^^:.=  1875. 
4 


By  the  rule, 


=  fV 


whence,  a;  =  100. 

12.  Two  persons,  A  and  B,  start  from  two  different  points,  and 
travel  toward  each  other.  When  they  meet,  it  appears  that  A  has 
travelled  30  miles  more  than  B.  It  also  appears  that  it  will  take  A 
4  days  to  travel  the  road  that  B  had  come,  and  B  9  days  to  travel 


175.]  EQUATIONS  OF  THK  SECOND  DEGREE.  151 

the  road  which  A  had  come.     What  was  their  distance  apart  when 
they  set  out  1 

Let  X  denote  the  number  of  miles  B  travelled; 

then  will     x  +  SO     "  "  "      A 


X 

4 


''      A  travels  per  day ; 


9 
^  +  30    , 


(I) 


^ 


{-V-l 


days  A         " 


«     B 


From  the  conditions, 

X  x  +  30  a;2       a;2  +  60^;  +  900 


or 


whence,  by  reduction, 

a;2  _  48a;  =  720  • 
and  by  the  rule, 

a;  =  24  ±  -^WO  +  576  =  24  d=  36 ; 
taking  the  upper  sign,         x  =  60,     and     a;  +  30  =  90  ; 
hence,  the  distance  is  150  miles, 

EXAMPLES  INVOLVING  RADICALS  OF  THE  SECOND  DEGRER 


(Z  I  <Z^  ——  iT*^  X 

3.  Given         -  +  \  / -7, —  =  r'    to  find  the  values  of  x. 

X        \       x^  0 


152  KEY  TO  DAVIES'  BOURDON.  [175. 

Multiplying  botn  members  by  bx,  and  transposing, 

b  -y/a^  —  a;2  z=  x"^  —  ab  ; 
•quaring  both  members, 

b^a^  —  b-'z^  -X*  -■  2abx^  +  a^^  ; 
cancelling  b'^a^,  dividing  both  menr.bers  by  cc^  and  transposing, 
x'^  =  2ab  —  b^         .-.         X  =  ±  •v/2r/6  —  fc^. 

4.  Given        \/-^^  +  2  \/—r—  =  b^  \/-^~,    to  find  x 
V        a;  Va:  +  a  Va;-fa 

Multiplying  both  members  by  \/~—~-j 

X  +  a        ^       fa 
X  V    X 

multiplying  both  members  by  a:,  and  transposing, 

2-v/ax  =  b'^x  —  X  —  a  =  {b"^  —  \)x  —  a-, 
squaring  both  members, 

Aax  =  (6*  -  2i2  4-  1)  a;2  _  2a  (^2  _  1)  x  +  a' ; 
transposing  and  reducing,  ' 

0:2  _     Mb'  +  1)  ^  ^   _ 


i*  -  262   -f-  1  ^4  _  262  _^   i' 

whence. 


a(b^  +  1) 


b*  -  262  _|_ 


_  _^_      I  a"  aHb*  +  26^  i-  1) 

1  ~  V       6*  -  262  -)-  1  "^   (64  _  262  ^  i\a' 


fl(62  +1)  2ab 


64  _  262  ^1   -  6*  —  262  +  1 ' 
now,  6*  -  262  -f  1  -  (6  _  ])2(6  ^  i)2_ 


175.]  PJQUATIONS  OF  THE    SECOND    DEGREE.  153 

Hence,  taking  the  upper  sign  and  reducing, 

_  a{h  4-  1)2  _         a{b  +1)2         _        a 


(62-1)2  (6-1)2(6  4-1)2  (6-1)2' 

ftnd  taking  the  lower  sign  and  reducing, 

a{b  -  1)2  a{b  -  1)2  a 


X  = 


(62    -   1)2  (b  -   1)2(6  +    1)2  ~  {b+    1)2' 

or,  uniting  the  two  values  in  a  single  expression, 

a 


X  =. 


(6+1) 


5.  Given,         —  =  b,     to  find  x. 

a  -\-  -y/  a2  —  a;2 

Clearing  of  fractions,  transposing,  &c., 


a(l -6)  =  (6  +  l)'v/o^-a;2; 

squaring  both  members, 

a2  (1  _  6)2  =  (6  +  1)2  (a2  -  x^) ; 
transposing  and  reducing, 

2  _      4£/26  2a  ^/b 

^.  V^+  1/2;  —  a  viP'a  -    , 

6.  Given,         -^^-= ^  .  — ,     to  find  x. 

y/x  —  -y/ar  —  a       X  —  a 

Multiplying  both  terms  of  the  first  member  by    -y/z  —  \/x  —  a, 
and  then  dividing  both  members  by    a, 

.        '  1  n2 


2ar  —  a  —  2  y/.c2  —  ax       x  —  a 
clearing  of  fractions,  &c., 


154  KEY  TO  DAVIES'  BOFKDON.  [175. 


(1  —  2»2)x  —  (1  —  »2)a  =  —  2«2  ^x^  —  ai  ; 

squaring  both  members,  transposing  and  reducing, 

2(1   -3;i2)a  1  _2«2  +  n*    , 

x^ ^ —  X  =^ a^  : 

1  -  4^*2  1  -  4ft2  ' 

whence  by  the  rule, 

_  1  —  3n2  /      (1  —  2w2  4-  n*)  a^         {\  —  Qn^  +  9n*)a2 

^  ~  1  -  47*2  <^  =^  V  V-^n^  '  (1  _  47i2p 

(1  -  3n2)  2w3a  (1  -  3^2  d=  2n^)  a 

1  -  4n2  1  -  4ra2  1  _  4/^2 

Taking  the  upper  sign,  and  dividing  both  terms  of  the  fraction  by 

1  +  2n, 

(1  —  2»  +  n^y        (1  —  nfa 

X  = —  =  -^ • 

1  -2n  1  -2n 

Taking  the  lower  sign,  and  dividing  both  terms  by   1  —  2», 

a(l  +  2ft  +  n2)        (1  +  ,,)2a 


X  = 


taking  the  two  values  together,  x 


1  +2ri  1  +  2n    * 


1  ±  2n 


7.  Given  — ::=  +     -— =\/ r-,  to  find  a;. 

^x  ^      V5 

Multiplying  both  members  by  V«, 
»quaring  ooth  members, 


2a  +  2  ya2  -  a;2  ^  —  .     ^j.,     2-v/a2  -  x2  =  y  -  '-i« ; 


175.]  EQUATIONS  OF  THE  SECOND  DEGREE.  155 

squaring  both  members, 

4  (a2  —  a;2)  =  — 1-  4a^  ; 

oancelling  4a^  in  the  two  members,  and  dividing  both  by  x^^ 

4a 


^    ~  ^-2  ^    ' 


clearing  of  fractions,  &c,, 


x"^  z=4ab  —  4b^  .'.  x  =  ±2  ^faF 


IP-. 


a  +  a;  +  ^/2ax  +  x^        ,      .     c  a 

8.  Given =  o  :  to  find  x. 

a  +  fr  ' 

Clearing  of  fractions,  transposing  and  factoring, 


■y/2ax  +  x'  =  {b—l){a  +  x). 
Squaring  both  members, 

2ax  +  a;2  =  (i^  _  26  +  1)  {a^  +  2ax  +  x^) ;     or, 
2ax  -\-x^  =  {IP  —  2b)  (a2  +  2ax  +  x^)  +  a^  +  2ax  +  «= ; 
whence,  by  reduction, 

whence,  by  the  rule, 


V  \     62  -  26     ./  ^6  _  fca 

a  (I  --/26~-62) 

taking  the  upper  sign,       x  = 7=7= -—^=  ; 

^26  —  6'' 

,.     .   ,        .  «(i  +-v/26^::>) 

takmg  the  lower  sign,       x  = : —         ; 

^  "  ^26  -  62       ' 

,"                                               ^a(l+v/26-6'« 
whence,  x  =  ± — ■' 

/26  -  6^ 


156  KEY  TO  DAVIES"  BOURDON.  [175-178. 

2d  Solution. 
Make  x  -\-  a  =  y,  whence,  2ax  +  a:'-  =  y*  —  a' 

substituting  in  the  equation,  and  clearing  of  fractions, 

y  +  Vy^  —  a^  z=by; 
transposing,  &c., 

yy2  _  a2  _  (5  _  1)  y . 

squaring  both  members,  and  cancelling, 

-a^  =  {b-'  -  26)  y2  ; 
whence,  solving  with  respect  to  y, 

a 

y=  ± 


^2h  -  \? 
substituting  for  y  its  value,  &c.. 


a 

a:  rr  —  a  d= 


y/26  -  62  ' 


whence,  as  before,         r  =_  —  -+-  V 


^26  -  U' 

TRINOMIAL    EQUATIONS. 

6.  Given         z*  -  (26c  +  4^2)  a-^  =  —  iV  ;  to  ftxi*/  x. 
By  the  rule. 


«  =  ±  ^hc  +  2a2  ±  y4o26c  +  4a*  =  dtyfic  +  2a2  ±  2a  y^  j    i'. 

7.  Given         2a;  -  7  y^  =  99  ;     or,     2a;  -  99  =  7  .^/r 
Squaring  both  members 

4«2  _  39Ga:  -1-9801  =  49x  : 


178-181]  EQUATIONS  JF  THE  SECOND  DEGEKE.  157 

transposing  and  reducing. 

,      445  9801 

4  4 

whence,  by  the  rule, 


445  /      9801 


or,  X 


198025 
~64~' 

445       203 


8    ~    8 
648 


taking  the  upper  sign,         x  =  — -  =  81 

o 

242       121 

"       lower  sign,         z  =  -— -  =  — — 

o  4 


d  c 

8    Given,  -  —  bx*  -\-  -x^  =  0,     to  find  x ; 


transposing  and  reducing. 

c  a 

X* x^  =  — 

bd  ^2 


=  ±\/^±J-  + 
V  2bd      \    b^^ 


C2 


wnence,  x-  ^  y  ^^^  ^  y   ^^^  ^^^^^  , 

reduchg,  x  =  ±  J  '  ^  Vi^^ 

^  V  'Zbd 


15  SAMPLES  OF  REDUCTION  OF  EXPRESSIONS  OF  THE  FORM  OF 


■\/«  ±  Vb. 


4.  Reduce  to  its  simplest  form,   \/28  +  10  ^3. 
o  =  28,     6  =  3Q0,     c  =  22. 


158  KEY  TO  DAVIES'  BOURDON.  TlSl. 

\pp]}  ing  the  formula  and  considering  only  the  upper  sign, 
x/284  10  ^'3  =  5  4-y^ 

5,  Reduce  to  its  simplest  form,    4  /l  +  4y—  3. 

a  =  I.  5=—  48,         c  =  7; 

applying  the  formula,  &c.. 


w  1  +  W-  3  =  24-  -/^^^ 

6.  Reduce  to  its  simolest  form, 


K/bc  +  n-^/bc'—  62  -  K/bc  -  nyjbc  -b^. 

a  =  bc,         b  =  462  (jc  —  62)^         c  =  6  (c  -  26) ; 
applying  the  formula  to  the  1st  radical, 

yJbc  +  26-v/6c  -  62  =  ±  (y/bc  -  62  +  6) ; 
applying  the  formula  to  the  2d  radical, 


^Jbc  -2b^bc  —  b^=  db(-v/6c-62-6); 
subtracting  the  second  result  from  the  first. 


y/bc-\-  26  ^bc  -  62  _y^ic  -  26  -/bi^^^  =  ±24. 
"i    Reduce  to  its  simplest  form, 


«/a6  -r-  4c'  -  rf2  _  2  y4a6c2  —  06^2. 
a  =  a6  +  4(;2  _  (^2^       6  =  16a6c2  -  4abcP,       c  =  ab  —  4c^   f-  <f-. 


181-183.]     EQUATIONS    OF    THE    SECOND    DEGREE.  159 

Applying  the  formula, 

^ab  +  4c2  —  (?!  —  2  V^abc^—ab(f=  ±  {Vab  —  V4:C^  —  d^)- 

SIMULTANEOUS  EQUATIONS. 

o  ■  {x  +  2y  =  7 (1) 

iz^+dxy  -\-y^  =  3l (2) 

From  (1), 

x  =  7  —  2y. 
Substituting  in  (2), 

49  _  282/  +  %'  +  3i/  (7  -  ^)  +  2/'  =  31 ; 

hence,  

7   ,    .749   ,    18  7_^11 

2^=-2±^T  +  T=-2±T' 

.-.      y'=2        and        y"=—9; 
and 

x'=d        and        a;"  =25. 

(2a: +  2/ =  27 (1) 

(3a;^  =  210 (2) 

From  (1), 

y  =  27  —  2x. 
Substituting  in  (2), 

Sx  (27  —  2x)  =  210 ; 
hence. 


27   .   a/T^V     o.  ___27±13 


-^i±V{T)-''=      4 


.-.    a;'=10        and        x"=d^. 
From  (1) 

y'z=7        and        y"=  20. 

^  j  2a:  -  3?/  -  1  =  0 (1) 

(2a;2+a:y-5t/2  =  20 (2) 

From  (1), 

"*-       2       ' 


IGO  KEY    TO    DAVIES'     BOURDON.  [183-184. 

from  (2), 


hence,  

7       ,/49       313       —7  ±19 
2^=-4±^16  +  l6"  =  -^r~' 

.-.     y'-Z        and        3/"=  -6^; 

and  from  (5), 

a;'=  5        and        «"=  —  9^. 

.    )^- •  •  ^^) 

(2/-^  =  2 (2) 

From  (2), 

2/  =  2  +  a;. 

Clearing  and  substituting  in  (1), 

lOx  +  2  +  a;  =  3a;  (2  +  a;) ; 

hence,  

_5         /25       24_5±7. 
*""6^*'^36'^36"~      6     ' 

a;'=  2        and        a;"=  —  ^. 


•    • 


From  (2), 

y=4        and        2/  =3 

(a;2-a;?/=    6 (1) 

^a;2+a;z/  =  66 (2) 

Adding  (1)  and  (2)  and  dividing  by  2, 

a;2=36;        .-.     a;'=  6        and        o(^'=—%\ 
hence, 

y'z=.  5        and        y"=  —  5. 

g_  ja:2-a;t/  =  48 (1) 


\7?  —  xyz^ 


xy-y^=12 (2) 


184.]  EQUATIONS    OF    THE    SECOND    DEGREE.  161 

Making  y=px, 

48 
x^-px^  =  ^^-     .'.    x^  =  -^^,    ...      (3) 

1  — p  ^ 

19 

px^-p2z2=12;    .'.     x^  =  -±±-       ...      (4) 

Equating  (3)  and  (4)  and  dividing  by  12, 


^2      "  1  5   ,    ./25       16 


hence, 

5 

4^~~4'     •  •    ^-8^  ^   64~6i 

or,  p  =  2    and    P  =  j' 

Using,    p  =  j,    we  have  from  (3), 

x'z=  +  8    and    «"=  —8;    whence,    y'=  +  2    and    y"=  —2. 

9.  ja;2  +  4;r2/  + 4^/2  =  256 (1) 

i3f  —  x^  =  39 (2) 

Making 

y=px, 
a;2  +  4:px^  4.  4jtj2.^2  =  256 (3) 

3p^x^  —  x^  =  39 (4) 

From  (3)  and  (4), 

3_          256 
^  ~  1  +  4j9  +  4j92 (^) 

39 

^  =  PTi («' 

Equating, 

256  _      39 

1  +  4^  +  4j!>2— 3^3_i5 

hence, 

768^2  _  256  =  39  +  156jo  +  156^^2, 

.                                    ,      13         295 
and  n2 n  = ! 

^       5V       612 ' 


163  KEY    TO    DAVIES'    BOUKDON.  [184. 

hence, 

78  ±432  5         J  59 

^  =  -613-'  "'  ^  =  6  "^^  p  =  -m- 

Using  the  first  value,  we  have  from  (6), 

a;2  =  36;  •.        «'=  6        and        a;"=— 6; 

and  by  substitution, 

y'=:  5        and        y"=.  —  5. 

Using  the  second  value,  we  find, 

ic'=102        and        a;"=— 102; 

and  by  substitution, 

y'=  —  59        and        y"z=  59. 

in  3  6  (xM- 2/')  =  13xy (1) 

\x^-f^=.20 (2) 

Making 

y=px, 

6(1  +^2)2;2_i3pa;2 (3) 

{l-f)x''  =  %0 (4) 

From  (3),  we  have, 

6  +  6p2  =  13;?. 

Reducing, 

_  1^       ./169       144  _  13  ±5 

■'•    ^~U^^  144       144  ~      12     ' 

or 

2 

p  =  l^        and       P  —  ^' 
From  (4)  we  have. 


20 

1— J9'^ 


a^  =  ^^ (5) 


188.]  EQUATIONS    OF    THE    SECOND    DEGREE.  163 

Using  the  second  value  of  p,  we  have  from  (5), 

20 

x^  — 7  =  36,     or    x'=6    and    x"=  —  6 ; 

1— i 

and  by  substitution,     «/'=  4    and    y"= — 4. 

EQUATIOJ^S  OF  A  HIGHER  DEGREE  THAN  THE  FIRST,  INVOLVING 
MORE  THAN  ONE  UNKNOWN  QUANTITY. 

ix^y   -{-  xv^   =    Q  .  .  .  (1)> 

15.  Given,  •{  "  M-    to  find  x  and  y.    • 

Dividing  (2)  by  (1),  member  by  member, 
xy  =  2,     or     x  =  — 

y 

Substituting  this  vahie  of  x  in  (1)  and  reducing, 

*  +  %  =  «; 

clearing  of  fractions  and  reducing, 

y2  _  3y  _  _  2 : 

u  3    ,      /     „  ,9        3        1 

whence,  y  =  -  ±^ -2  +  -  =  -  ±  j; 

or,  y  =  2,     and     y  =1; 

whence  x  =  \,     and     x  =2. 

16.  Given,      ]  ^      ^  ^  ^  ^  to  find  a?  and  y. 

(ary  =6 (2)  )  ^ 

Multiplying  both  members  of  (2)  by  2  and  adding  the  resulting 
equation  to  (1),  member  to  member, 

a;2  +  2xy  +  y2  4-  J;   f  y  =  30, 
or,  {x   4- y)2  +  a:4-y  =  30; 


164  KKY  TO  DAVIES'  BOURDON.  [189. 

whence,  by  tlie  rule, 

whence,  x  -\-  y  =  5,     and     x  -\-  7/  =  —  Q. 

Taking  the  first  value  of  x  +  y  and  substituting  in  it,  for  y  Its 
value  —  derived  from  (2), 

X 

6 
X  -{ =  5  ; 

X 

clearing  of  fractions  and  reducing, 

x^  —  5x  =  —  6  ; 


5  /      ^    ,    25       5         1 

whence,  x  =  — ±^^-6+— =  -±-; 

iyr,         ar  =  3,     and     a;  =  2 ;     whence,     y  =  2,     and     y  =  3. 
Taking  the  second  value  of  x  +  y  and  proceeding  as  before, 

x-\ =  —  6  ; 

X 

clearing  of  fractions,  die, 

x^  ■\-  Gx  =  —  6 ; 

whence,  z  —  —S  ±  y/—  6  +  9  =  —  3  db  /3 ; 

and  by  substitution,  y  =  —  3  qp-\/3. 

PROBLEMS  GIVING  RISE  TO  EQUATIONS  OF  A  HIGHER  DEGREE 
THAN  THE  FIRST  CONTAINING  MORE  THAN  ONE  UNKNOWN 
QUANTITY. 

2.  To  find  four  numbers,  such  that  the  sum  of  the  first  and  fourth 
shall  be  equal  to  2s.,  the  sum  of  the  second  and  third  equal  to  2s', 
the  sum  of  their  squares  equal  to  4c"'',  and  the  product  of  the  first 
*nd  fourth  equal  to  the  product  of  the  second  and  third. 


189-191.  J      EQUAi:oNS  of  the  second  degree,  165 

Assuming  the  equations, 

u  +  z  =  2s     •     •     •     •     (1) 
a;  +  y  =  2s'    .     .     .     '     (2) 

tt2  +  X2  +  y2   4.    22   _  4c2      .        .        .        .        (3) 

uz  =  xy     •     •     •     •     (4) 

Multiplying  both  members  of  (4)  by  2,  and  subtracting  from  (3), 
member  from  member, 

tt2  —  2uz  +  22  +  a;2  +  2xy  +  y^  =  4c^ ; 

or,  *  (w  —  z)2  +  (a;  4-  t/)2  z=  4c2, 

Substituting  for  a;  +  y  its  value  2s'  and  transposing, 

{u  —  2)2  =  4c2  —  4s'2, 


or,  u  —   z  =  -^402  —  4s'2. 

Combining  with  (1), 

,    -i/4c2  -  4/2  ^ 

« =  s  +  -^^ — ~ —  -  s  +  -v/c'  -  «'^ 

■v/4c2  —  4s'2 


and,  2  =  s =z  s  —  -yjc^  —  «'2  ; 

reversing  the  order  of  the  members  of  (4),  and  proceeding  as  before, 
we  find  in  like  manner. 

a;  =  s'  4-  -y/d^^^—^^ 


and.  y  =  s'  —  ^c^  —  s^. 

4.  The  sum  of  the  squares  of  two  numbers  is  expressed  by  i 
and  the  difference  of  their  squares  by  b  :  what  are  the  numbers  1  , 

Let  X  and  y  denote  the  numbers. 

From  the  coi  ditions  of  the  problem, 


166  KEY  TO  DAVIES'  BOURDON.  [191. 

x^  -\-  y^  =  a     •     •     •     •     (1) 
x^  -  y^  =  b     .     .      .     .      (2). 

By  adding,  member  to  member, 

2x^  =  a  +  b  .  • .  X 

by  subtracting, 

2y2  =  a  —  b  .  ' .  y 

5.  What  three  numbers  are  they,  which,  multiplied  two  and  twu 
and  each  product  divided  by  the  third  number,  give  the  quotients 
a^b^cl 

Let  a:,  y  and  2,  denote  the  numbers 
From  the  conditions  of  the  problem, 


=• 

/a  +  b 
^V      2      ' 

^.A-* 

xy 


z 


y^ 


■=  a     or,     xy  =^  az     •     •     •     (1) 


—  =  b     or,     yz  =  bx     •     •     •     (2) 


z 


xz 


—  =  c     or,     xz  z=  cy     •     •     •     (3). 

y 

Multiplying  (1),  (2)  and  (3)  together,  member  by  member, 

x'^yH'^  _  abcxyz ; 

dividing  both  memoers  by  xyz, 

xyz  =  abc     •     •     •     (4) ; 

gubstituting  in  (4)  the  value  of  xy  taken  from  (1),  and  dividing  both 
members  by  a, 

z^  =  be  .  • .  z  =z  ^bc. 

Substituting  the  value  of  yz  and  dividing  by  6, 


191.]  EQUATIONS  OF  THE  SECOND  DEGKEE.  167 


X'^  =  ac  .  • .  «  =  y  «C. 

Substituting  the  value  of  xz  and  dividing  both  members  by  c, 

6.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their  cubes  is 
152  ;  what  are  the  numbers'? 

Let  X  and  y  denote  the  numbers. 

From  the  conditions, 

x-\-y=      8 (1) 

x3  +  y3  _  152 (2)  ; 

cubing  both  members  of  (1), 

a:^  -f  ^x'y  +  3xy2  +  y^  =  512     •     •     •     •     (3) ; 

subtracting  (2)  from  (3),  member  from  member,  and  dividing  both 
members  by    3, 

a:2y  +  ary2  =  120     •     •     •     •     (4)  ; 
substituting  the  value  of  ar  taken  from  (1), 

(64  -  16y  +  y2)  y  4.  (8  _  y)  ^2  ^  120, 
or,  reducing,  y2  _  gy  ~  —  15 ; 


whence,      y  =  4  ±  -y/—  15  +  16  =  4  ±  1     .  • .     y  =  5,     y  =  3 ; 
whence,  from  (1)  a:  =  3,     a:  =;  5. 

7.  Find  two  numbers,  whose  difference  added  to  the  difference  :( 
their  squares  is  150,  and  whose  sum  added  to  the  sum  cf  theii 
squares,  is  330. 

Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 


168  KEY  TO  DAVIES'  BOUKDON.  [191. 

X^-y2  +  X-y  =lbO      .      .      .      .      (1), 

x^ -{- y' -\-  X  +  y  =z  Sm     .     .     .     .     (2) ; 
adding  member  to  member,  and  reducing, 

a:2  +  a;  =  240 ; 


whence,  ar  =  -  -  ±  -/240  -f  -  =  -  -  d=  _ ; 

or,  considering  only  the  positive  solution, 

X  =\5; 
vhence,  from   (1),  by  substitution, 

8.  There  are  two  numbers  whose  difference  is  15,  and  hf.lf  their 
product  is  equal  to  the  cube  of  the  lesser  number :  what  are  the 
numbers  1 

Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

X  —  y  =  \5      •     .     .     (1), 

^  =  y3^     or     a:=:2y2     .     .     .     (2); 

(substituting  in  (1)  and  dividing  both  members  by  2, 

y     15 
y"  — 

wlience. 


y2   _  £.  _ 

2        2 


r    o  -  o » 


1  /15        1         1       11 

y=4--^V2-+16  =  4=^T 


considering  only  the  positive  solution, 

y  =z  S  \     whence,  from  (1),     x  =  18. 


191-192.]     EQUATIONS  OF  THE  SE'::;OND  DEGRER.  ICO 

9.  What  two  numbers  are  those  whose  sum  multiplied  bj  the 
greater,  is  equal  to  77  ;  ai.d  whose  difference,  multiplied  by  the 
lesser,  is  equal  to   12? 

Let  X  and  y  denote  the  numbers. 

From  the  conditions, 

{x  -{-  y)  X  -  77,  or     x"^  +  xy  -^11     -     •     •     (1)  ; 

{x  —  y)y=\2,  or     xy  —  y"^  =  \2     •     •     •     (2); 

make     x  ■=  py  ;     whence, 

77 
{p-^-^p)y'^  =  ll,     or,     y^=-^—-     •     .     .     (3), 

12 

{p  -1)3/2  =  12,    or,    y^  =  -r^^  •    •    •    (4); 

equating  the  second  members  and  reducing, 


65  77 

^^-12^=-T25 


n2 


,  65  / 77   ,   4225       65       23 

whence,         »  =  —  ±.\,  /  —  —  A =z  —  ±  — ; 

-^24      V        12  ^  576        24       24' 

taking  the  upper  sign, 

_88_11_ 
^~24~"3"' 

,     .      .       .     /  .V  /36       3      /— • 

subslitutmg  u    (4),         y=\/— =  -y2j 

whence,  a:  =  —  ■y/2; 

42       21 

taking  the  lowei  sijrn  »  =  — -  =  — - ; 

»  b  ^      24       12* 


170  KEY  TO  DAVIES'  BOURDON.  [192. 

substituting  in  (4), 

y=z\J-—-z^A\      whence,      ar  =  7. 

10.  Divide  100  into  two  such  parts,  that  the  sum  of  the.f  squart 
roots  may  be  14. 

Let  X  and  y  denote  the  parts. 

From  the  conditions, 

x+      y  =  100     .     .     .     (1), 

-v/^+  •v/y=     14     ...     (2): 

squaring  both   members  of  (2)  and  subtracting   (1),  member  from 
member. 

2^/xy  z=z  96,     or     -\Ay  =  48,     or     xy  =  2304  : 

substituting  for  y  its  value,  100  —  x, 

100a;  -  x^  =  2304,     or     x^  —  \00x  =  ~  2304. 

■whence,  by  the  rule. 


x  =  50db-/196r=50  ±  14; 
hence,  a;  =  64,     x  —  36, 

and  y  =  36,     y  =  64. 

11.  It  is  required  to  divide  the  number  24  into  two  such  parts, 
that  their  product  may  be  equal  to  35  times  their  difference. 

Let     X  and  y     denote  the  parts. 
From  the  conditions  of  the  problem. 

a;  +  y  =  24     .     •     .     (1) 
xy  =  35{x-y)     .     .     .     (2); 
substituting  in   (2)   the  value,     y  z=  24  —  ar, 

24ar  -  a;^  =  35  (2x  -  24),     or,     24x  -  x^  =  70.r  -  840 ; 


192.]  EQUATIONS  OF  THE  SECOND  DEGKEE.  171 

wheivce,  a;*  +  AQx  =  840  ; 

by  the  rule, 

«  =  -  23  ±  ^840  +  529  =  —  23  ih  37  ; 

ence,  taking  the  upper  sign,     a;  =z  14 ; 
Dy  substitution,     in   (1),  y  =  10. 

12,  What  two  numbers  are  those,  whose  product  is  255,  and  the 
sum  of  whose  square  is  5141 

Let  X  and  y  denote  the  numbers. 

From  the  conditions,         xy  =  255     •     •     .      (1) 

x^  +  y^  =  514  .     .     (2), 

multiplying  both  members  of  (1)  by  2,  adding  and  subtracting  the 
resulting  equation  to  and  from  (2),  member  by  member, 

a;2  +  2x?/ +  y^  =  1024     .     .     (3) 
x^  —  2x1/  +  2/2  _  4      ^     ^     .     (4) ; 

extracting  the  square  root  of  both  members, 

a:  +  y  =  32, 

x-y  =  2, 
whence,  a:  =  17 ;         y  =  15. 

13.  There  is  a  number  expressed  by  two  digits,  which,  when 
divided  by  the  sum  of  the  digits,  gives  a  quotient  greater  by  2  than 
the  first  digit;  but  if  the  digits  be  inverted,  and  the  resulting  num- 
ber be  divided  by  a  number  greater  by  1  than  the  sum  of  the 
digits,  the  quotient  will  exceed  the  former  quotient  by  2 :  what  is 
the  number  1 


173  KEY  TO  DAVIES'  BOUKDON.  [1S2. 

Let  %  and  y  denote  the  digits ;  then  will 

lOx  +  y  ienote  the  number. 
From  the  conditions, 

i^^  =  x  +  2    .     .     (,) 

lOy  +  a: 


=  a:  +  4     .  (2); 


2:  +  y+  1 

clearing  of  fractions,  and  reducing, 

%x  —  y    —  a;2  —  a-y  =  0     •     •     •     (3), 
6y  —  4.r  —  a;2  —  ary  =  4     •     •     •     (4) ; 

by  subtraction, 

7y-12x  =  4;         .'.     y  = ^ ; 

substituting  in  (3), 

12a;  +  4         ,       12a;2  +  4a;       ^ 
8x ^ a;2 ^ —  =  0, 

7  7  ' 

clearing  of  fractions  and  reducing. 


a;2  —  —  X  =  —  — ,     whence, 

20  I      4       400      20       18 

x=.  —  ±\/ 1 =  —  db  — • 

19      V        19^301       19       19 

Taking  the  upper  sign,  gives     a;  =  2 ;     whence,     y  =  4. 

14.  A  regiment,  in  garrison,  consisting  of  a  certain  number  of 
companies,  receives  orders  to  send  216  men  on  duty,  each  company 
to  furnish  an  equal  number.  Before  the  order  was  executed,  three 
of  the  companies  were  sent  on  another  service,  and  it  was  then 
found  that  each  company  that  remainod  would  have  to  send  12  men 
additional,  in  order  to  make  up  the  complement,  216.     How  many 


192.]  EQUATIONS  OF  THE  SECOND    DEGREE.  173 

companies  were  in  the  regiment,  and  what  number  of  men  did  each 
of  the  remaining  companies  send  1 

Let  X  denote  the  number  of  coznpanies,  and 

y       "        "         "  each  should  send ;  then, 

y  +  12     will  denote  the  number  sent  by  each. 

From  the  conditions  of  the  problem, 

xr/  =  21Q     •     •     •     (1), 

{x  -  3)   (y  +  12)  =  216     .     .     .     (2)  ; 

Performing  operations,  subtracting  and  reducing, 

4ar  —  y  =  12.  .  * .       y  =  4a;  —  12 ; 

substituting  in  (1),      Ax^  —  12.r  z=  216,     or     a;^  —  Sa;  =:  54 ; 


whence,  a;  =  —  ± 'y/54  +  -  =  —  ±  —  ; 

taking  the  upper  sign, 

'  a;  =  9  ;         hence,         y  =  24,     and     y  +  12  =  36. 

15.  Find  three  numbers  such,  that  their  sum  shall  be  14,  the 
sum  of  their  squares  equal  to  84,  and  the  product  of  the  first  and 
third  equal  to  the  square  of  the  second. 

Let  X,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

a;   +  y    +  z    =  14     •     •     •     (1), 

a;2  +  y2  +  22  _  84     .     .     .     (2), 
xz  =z  y"^      •     •     '     (3). 
Multiplying  both  members  of  (3),  by  2,  adding  to  (2)  and  'educing, 


174  KEl  TO  DAVIES'  BOURDON.  [192. 

^   x^+2xz  +'22  =  84  4-  y2 .  .  _     x  +  z=  ^b4  -j-  y^  •  •  (4  j 

from  (1),  a;  +  2  =  14  —  y  •  •  .  .  (5); 

equating  the  second  members  of  (4)  and  (5)  and  squaring, 

84  + y2  =  196  —  28?/ +  y2;         ...     y  -  4. 
Substituting  in  (1)  and  (3), 

a;  +  2  =  10     .     .     .     .     (6) 

ar2  =  1 6     .  • .     X  =  — . 

z 

Substituting  in  (6)  and  reducing, 

22-102  =   -  16 

2  =  8;         2  =  2, 

and  by  substitution,        x  =z  2 ;         a;  =  8, 

16.  It  is  required  to  find  a  number,  expressed  by  three  digits, 
such,  that  the  sum  of  the  squares  of  the  digits  shall  be  104 ;  the 
square  of  the  middle  digit  to  exceed  twice  the  product  of  the  other 
two  by  4  ;  and  if  594  be  subtracted  from  the  number,  the  remainder 
will  be  expressed  by  the  same  figures,  but  with  the  extreme  digits 
reversed. 

Let  X,  y  and  z  denote  the  digits ; 

then,  100a:  +  \Qy  +  z    will  denote  the  number. 

From  the  conditions  of  the  problem, 

a;2  +  y2_|_22_  104     .     .     .  (1) 

y2_2x2  =  4         ...  (2) 

100a:  +  lOy +2  —  594  =1002  +  lOy +  a;  (3); 


193.]  EQUATIONS  OF  rilE  SECOND  DEGREE.  175 

subtracting  (2)  from  (1),  member  from  member, 

x2  -I  2x«  +  22  =  100         .  • .         X  +  z  =  10; 
reducing  (3)  a;  —  z  =  6  ; 

hence,  x  =:  8,     and     2=2. 

By  substitution,        y  =  6,     and  the  number  is  862. 

17.  A  person  has  three  kinds  of  goods  which  together  cost 
$230j*^.  A  pound  of  each  article  costs  as  many  ^  dollars  as  there 
are  pounds  in  that  article  :  he  has  one-third  more  of  the  second  than 
of  the  first,  and  3i  times  as  much  of  the  third  as  of  the  second . 
How  many  pounds  has  he  of  each  article  1 

Let  Xf  y  and  z  denote  the  number  of  pounds  of  each  article.    N's- 

From  the  conditions  of  the  problem, 


a;2     .    y2        ^2        5525 

+        4-        =          'or, 
24   '   24   '   24         24 

a;2  +  y2  ^_  ^2  ^ 

=  5525     (1) 

4 

t='P  ■ 

•     .     (2) 

7                           14 

2       196    „ 

.2  =    ^a- 

.     .     .    (3) 

substituting  in  (I)  and  reducing, 

a:2  =  225 

a;  =  15, 

substJtu'.ing  in  (2)  and   (3), 

y  =  20; 

«=70. 

18.  Two  merchants  each  sold  the  same  kind  of  stuff:  the  seccnd 
sold  3  yards  more  of  it  than  the  first,  and  together,  they  received 
35  dollars.  The  first  said  to  the  second,  "  I  would  have  received  24 
dollars  for  your  stuff."     The  other  replied,  "  And  I  would  have 

7 


176  KEY  TO  DAVIES'  BOURDON.  [193. 

received   12^  dollars  for  yours."     How  many  yards  did  each  of 
thorn  sell  1 

Let  X  and  y  denote  the  number  of  yards  sold  by  each. 

24 

Then  will  —  denote  the  price  the  first  received  per  yard, 

y 

25 

nn^  —  -vv-ill  denote  the  price  the  second  received  per  yard. 

From  the  conditions  of  the  problem, 

X  +  'S  =z  7/, 
?1^  +  ?^  =  35,     or,     48x2  ^  25y2  =  lOxy  : 

y        2x 

substituting  in  the  second  equation  the  value  of  y  taken  from  the 

first, 

48x2  +  25  (x2  +  6x  +  9)  =  70x2  _|_  210x ; 


reducing. 

x2  -  20x  =  -  75  ; 

whence. 

x   =  10    ±  y/25  =  10  ±  5, 

or, 

ar   =  15  ;             x  =  5  ; 

substituting, 

y  =  18  ;            y  =  S. 

19.  A  widow  possessed  13000  dollars,  which  she  divided  into 
two  parts,  and  placed  them  at  interest,  in  such  a  manner,  that  the 
inoomes  from  them  were  equal.  If  she  had  put  out  the  first  portion 
at  the  same  rale  as  the  second,  she  would  have  drawn  for  this  part 
360  dollars  interest ;  and  if  she  had  placed  the  second  out  at  the 
same  rate  as  the  first,  she  would  have  drawn  for  it  490  dollars 
interest.     What  were  the  two  rates  of  interest? 

Let  X  and  y  denote  the  rates  per  cent. 

Let  2  denote  the  1st  portion  ;  then  will  13000  —  z  denote  the  2d. 


193-201.]  BIXOMIAL    FORMULA.  177 

rrom  the  conditions  of  the  problem, 

xz        (13000  — z)«  „^^„ 

100  =  ^ rOO~     '  ''''         ^^=13000y-.y     .     •     •     (1), 

j^  =  360,  or,         zy  =  36000  •     •     •     (2), 

(13000  —  z)  X 

Y^ — ^  =  490,         or,         13000^ -0a;  =  49000  •     •     (3). 

Substituting  in  (1)  the  values  of  zy  and  zx  taken  from  (2)  and  (3) 
and  reducing,  we  find,  x  =  y  -\-  \. 

Substituting  this  value  of  x  and  the  value  of  z  taken  from  (2)    in 
(1),  and  reducing,  we  find 

2  _  72     _  36 
^   ~l3^  "~  13' 


36  /36       1296       iQ  ±  42 

whence,       y   =  —  ±\/ 1 =z ;        .  •       w  =  6  ' 

'       ^         13      V  13^    169  13       '        •    .     y  —  o, 

by  substitution, 

a?  =  7,     and     z  =  6000,     13000  -  z  =  7000. 

BINOMIAL    FORMULA. 

Art.  141,  I),  201. 

(5.) 
32a;5_j_  16^:4  ^     ^^^    ^       4.^.2    +     2a;    +     1 

1     +     3?/    +     9^/2    +     •Z'iif    +  ^lif  +243?/5 

1+5+10       +10       +5+1 
32a;5  +  240a:*?/  +  7202;y  +  1080.Ty  +  810a;^4  +  243?/5.   Ans. 

(6.) 

«s  +     a6      +     a"  +     fl2  ^1 

1    —     x^      -{-     x^  —     x^  +   x^ 

1+4  +6  +4  +1 

a^  —  ^a^x^  +   Qa'^x^  —  4:a^x^  +  a^.     Ans. 


178  KEY    TO    DAVIES'    BOURDON.  [201-207. 

1    +    1+1        +1+1        +1 

1   +   5     +10      +10      +5+1 

1   +  |i-  +  ix^  +  |a^  +  A^  +  ?V^^-    ^ns. 

(8.) 

1+1+1+1+  1+         1 

1   +     4a;  +     16a;2  +     64t3  ^  256a;4  +  10242^5 

1+5     +     10+10+  5+         1 

1  +  20a;  +   160a;2  +  640a;3  +  1280a:4  _^  I024a:5.  Ans. 

EXTRACTION    OF    HIGHER    ROOTS. 

Art.  144,  2>'  207. 

(2.) 
27^054^036^008  |  3002.    Ans. 

33  =  27 

3  X  3002  _  270000  I  054  036  0 


30023  =  27  054  036  008 

ExPLANATiox. — We  see  at  once  that  the  second  figure  of  the  root  is  0. 
Bringing  down  three  more  figures  and  forming  a  new  trial  divisor,  3  x  300^ 
we  see  that  the  third  figure  of  the  root  is  0.  We  then  bring  down  three 
more  figures  and  proceed  as  shown  above. 

(3.) 

483^249  |_78.    Ans. 
73  =  343 


3x72  =  147  1^40  2 
783  ^  474  552 


8  697.    Rem. 


207-208.]  EXTEACTION    OF    HIGHER    ROOTS.  179 

Note. — In  finding  the  second  figure,  we  see  that  9  is  too  large  and 
therefore  we  try  8. 

(4.) 

91^632^508^641  |  4508.  Ans. 
43  =  64 

3  X  42  =  48  I  27  6 


453  =  91 125 


3  X  4502  _  607500  i  507  508  6 


45083  =  91  611  864  512 

20  644  129.  Rem. 

(5.) 

32^977'340^218'432  |  32068.  Ans. 
33  =  27 
2  X  32  =  27  I  5  9 


323  _  32  768 


3  X  3202  =  307200  |  209  340  2 


32063  =  32  952  665  816 


3  X  32062  —  30835308  I  24  675  402  4 


320683  =  32  977  340  218  432 

,  (6.) 
2^197125  I  130.  Ans. 

13  =  1_ 

3x12  =  3  I  11 

133  =  2  197 


3  X 132  =  507  I  0  1 


1303  =  2  197  000 


125.  Mem. 


180  KEY    TO    DAVIES'    BOUKDON.  [208-213. 

(7.) 

1^035^678  |_101.    Ans. 
13  =  1 

3  X 102  =  300  I  35  6 


1013  =  1  030  301 


5  377.    Rem. 


Art.  144*,  p.  208, 

(2.) 
81^3165  j_30.    Ans. 
3^  =  81 


4x33  =  36  I  03 


304  _-  81  0000 

3165.     Rem. 

(3.) 
273^85964  |_30.    Ans. 
35  =  243 


4  X  34  =  324  I  30  8 


305  _  243  00000 


30  85964.     Rem. 


Art.  148,  j>.  213, 

(1.) 
_  473  X  8000  _  3784000 
~        8000        ~      8000 


213.]  EXTEACTIOK    OF    HIGHER    ROOTS.  181 

3^784^000  I  155 

13=   1_ 

3  X 12  =  3  \^^  ...  y^3  =  i^=z  7.75.  Ans. 
153  _  3  375 


3x152  =  675  I  409  0 
1553  =  3  723  875 

60 125.     Bern. 

(2.) 

79.'000'000'000'000  |  4.2908.    Ans. 
43  =  64_ 
3  X  42  =  48  I  iT^ 


423  _  74  088 


3  X  422  =  5292    4  912  0 


4293  =  78  953  589 


3  X  42902  =  55212300    46  411  000  0 


429083  =  78  997  767  077  312 

2  232  922  688.     Rem. 

3.  First  extract  the  square  root  to  six  decimal  places,  then  extract  the 
cube  root  of  the  result. 

FIRST   OPERATION. 

13.^00^00^00^00^00^00  I  3.605551 
9  


66  )  4  00 
3  96 


7205  )  4  00  00 
3  60  25 


72105  )  39  75  00 
36  05  25 


721105  )  3  69  75  00 
3  60  55  25 


7211101  )  9  19  75  00 
7  211101 


1  98  63  99.  Rem. 


182  KEY    TO    DAVIES'    BOUEDON".  [213. 


SECOND  OPERATION. 

3/605^551  I  1.53.    Ans. 
13  =  1 


3x12  = 

3    2 

6 

153 

=  3 

375 

3  X 152  = 

675 

230  5 

1533 

=  3 

581  577 

33  974.     Rem. 

(4.) 
3.004^150^000^000  |  1.4429.    Ans. 
13  =  1_ 

3  X 12  =r  3  I  20 

143  3=  2  744 


3  X 142  =  507  I  26  01 


1443  _  2  985  984 


3  X  1442  —  62208  |  1  816  60 
14423  _  2  998  442  888 


3  X  14422  —  62338092  I  570  711  20 


144293  _  3  004  060  675  589 


89  324  411.    Bern. 

(5.) 

O.OOI'OIO  IJ^    A71S. 
13  =  .001 


3  X  .12  =  .03  I  .000  01 


.103  =  .001  000 


.000  010.    Eem, 


213-217.]  EXTEACTION    OF    HIGHER    ROOTS. 


183 


(G.) 
^  =  .560^000^000  I  .824.    Ans. 
.83  =  .512 


3  X  .82  =  193    48  0 


.823  _  .551  368 


3  X  .822  _  20172    8  632  0 


.8243  _  .559  476  224 


.000  523  776.    Rem. 


Art,  150,  2)p.  217-218. 

(2.) 

a;6  +  6a:^— 40a;3^96a;— 64  i  a;2  +  2a:— 4.     Ans. 
(x^  +  'Zxf  =  a;6  +  G.i-5+12a;^  +  8a;3 

—  12a:^— 48.r3,  etc. 
(a;2+2a;— 4)3  =  a:6  +  6a:5— 40a^  +  96a;— 64 


ooc!^.     Dimsor. 


(3.) 

24a^,  etc.     

(2a;'-a;  +  2)3  =  8ic«-12a!5  +  30a^-25a^  +  30a;^-12ir  +  8 


2x^ — a; +  2.     Ans. 


12x*.     Divisor. 


4.  In  the  following  examples  we  first  extract  the  square  root  of  the 
given  polynomial,  and  then  the  square  root  of  the  result. 

FIRST   OPERATION. 

16a4_96a3a;  +  216a2a;2_216aa;3_,_8i2;4  jJ^—Uax  +  d^ 

16a* 

8a2_  12aa;  )  —  96a3a;  +  216a2:c2 
—  96a3a:  +  144a2a^ 


8a2— 24aa:  +  9a;2)  72a2a;2— 216ax3  4.8la:4 
72a2'c2_216fla;3  +  81a;* 


184  KEY    TO    DAVIES'    BOUEDOl^.  [217-218. 

SECOND  OPERATION. 

4«2  _  I2ax  +  9x^  i2a  —  3x.    Ans, 
4a2 


4,a  —  3x)—  12ax  +  9x^ 
—  12ax  +  9a;2 

0 

5.  In  this  and  the  following  example  we  first  extract  the  square 
root  of  the  given  polynomial  and  then  we  extract  the  square  root 
of  the  result. 

FIRST  OPERATION. 

a;8  _  4a;s  +  6x^  —  Ax'^  +  x'^  \  x*  —  2x  -\-  x'^ 


2x1^  —  2x  )  —  4:7^  +  6a;2 
—  4a^  +  42;2 


2a:4  —  4a;  +  a;-2  )  2x^  —  4a;-i  +  x'^ 
2a;2  —  4a;-i  +  x-^ 


0 


SECOND  OPERATION. 

xf^  —  2x  +  x-^\x^  —  ar\    Ans. 
x^ 


2a;2  —  x-^)—2x-\-  x'^ 
—  2x  +  x-"^ 

0 


218-220.]  TRANSFORMATION    OF    RADICALS.  185 

(6.) 

FIRST  OPEKATION. 

8l3f "  ■* 

I8ari-12x3)  -216a;' 

-216X'' +  U4afi 


18x4-24i;3_8a;i )  -lUx^  +  3diix^-56x* 
— 144x6 +  192a;=  +  64r* 


18a:*- 24^3-1 6a;' +  8a! )  144a;5_i20j;i_224a;» 

144a;5  _  I92a;^  - 1 282'^  +  64a;' 


18a;*-24a;3-16a;2  +  16j;  +  4  )  72ar*-  96a«'-64a;'^  +  64a;  +  16 

72a;*-  96a;3-64a;'  +  64a;  +  16 


0 

SECOND  OPERATION. 

9x*  —  12a:3  —  8x^  +  Sx  +  4:  |  3a^  —  2a:  —  2.     Ans. 
9«* 

6a;2  —  2x)—  I2x^  —  Sx^ 
—  Vla^  +  4a:2 


6a;2  —  4a;  —  2  )  —  12.^2  +  8a;  +  4 
—  Ux^  +  8a;  +  4 

0 

TRANSFORMATION    OF    RADICALS. 
Art.  152,  p2).  220-221. 

1.     V^SaWd^  =  \/l0aVxTf^2  _  2ac\/^ab^.    Ans. 


2.  V112a^  =  \^Sa^b^  x  14a  =  2aIPVl4:a.     Ans. 

3.  2^54a;-''y-6  =  2\^27x^y-^  x  2a;-^ 

=  ex-^y-^\^2x~K    Ans. 


186  -   KEY    TO    DA  VIES'    BOUEDOJST.:  i"  [220-222. 


*  5.     3Vl28a.-5j/-5  =  3v^64^V^x2':?;2t/-2 


=  12zj/~^y2x^i/~^.    Ans. 


6.  V  —  24:a;-y  =  ^_  s.t-Si/S  x  3^/2  =  -  %x-^yy/2>if.  Ans. 

7.  S-v^Sa^  =  •v^27x2a^'  =  vb^ax.    Ans. 

8.  —  4\/aJ2  ^  ^(—4)4^"^  =  'C/^SGo^.    ^«s. 

9.  2ab\/2a^y^^  =  ^/lQa^¥x2ar^Y'^  =  ^^2abHj^.     Ans. 

10.     —  3x-^\/—  2y^  =  v'^I^z-^f  X  —  2f 

=  \^—21x-^  X  —  2f  =  \/5^x-Y.    Ans. 

Art.  153,  p.  222. 

1.     ^/^&aW  =  y  V^QaW  =  Vead.     Ans. 


2.     y^lGaiz^i/"^  =  \/ ^/iQa^x^y-'^  =  \/4:a^xy-i    Ans. 


3.  \^27m^ny  =  \/ \^21m^nY  =  V3m^)7^jj.    Ans. 

4.  v^"8a3^  =  y  y/'—  8a3my«  =  V—  2a^y\    Ans. 


5.     -^S-r^^^-S"  =  |/v^8a;3y-3n  =  ^2xy-^.    Ans. 


6.     v^l69*6c-4  =  y  ^/imb^c-^  =  \/lSb^c-\    Ans. 


223-224.]  ADDITION   AND   SUBTKACTION. 


18? 


Art.  154:,  p.  223. 

1.     The  1.  c.  m.  of  4,  6,  and  8,  is  24. 


24/ 


2.  The  1.  c.  m.  of  4,  6,  and  2,  is  12. 

^  =  ^27,     \/7  =  v^49,     V2  =  v'(2p  =  ^^04.    Ans. 

3.  The  1.  c.  m.  of  3,  6,  and  4,  is  12. 

2'V^  =  2v^8l^,     G\/Ix  =  6v^lto"^     A^a  =  V"^.    Jws. 

4.  The  1.  c.  ra.  of  3,  4,  and  6,  is  12. 
^^=^/7i^,     ^b^=^W^,     V^=^/7^.    Ans. 

5.  The  1.  c.  m.  of  m,  n,  and  p,  is  mnp. 


^a  =  T^,     ^b  =  "^^b^v,     ^c  =  "7c^^.    Ans. 

6.     The  1.  c.  m.  of  2  and  3,  is  6. 

1  fi  /       i 

Ans. 


a  —  b 


{a  —  bf      \  X  —  y 


{x  —  yY 


ADDITION  AND  SUBTRACTION  OF  RADICALS. 


1.  VJSaf^=4:bV^,    and    bV^f5a=5bVSa ;   .-.   Ans.QbV^a. 

2.  3  V4^  =  3V2a;    2  V^,  .-.   Ans.   bV^a. 

3.  2\/45  =  6\/5;    3^/5,  .-.   Ans.   rVsT 

4.  3a  V*  —  2c  V^,  .-.   Ans.  (3a  —  2c)  Vb. 

5.  3  V4a2  —  2  V3a ; 

3V4^  =  3V2a;  .-.   ^ws.    V^- 

6.  a/243  =  9  a/3  J  a/27  =  3a/3;  a/48  =  4  a/3.    Ans.  16  V^. 


188  KEY    TO    DAVIES'    BOUEDON.  [224-225. 


7.  2^2^  X  4a2  _  7a  V^a  x  9  +  5  V^a  x  SGa^  —  \/2a  x  25^-2 
=  (4a  —  21a  +  30a  —  5b)  V^a  =  (13a  —  5b)  V2a. 

8.  12  V2'x  1  +  3  Vr^  =  (6  +  I)  \/2  =  -^  V2. 


9.  V(2a  +  h)  8a3  -  V(2a  +  b)  b^  =  (2a  -  ^)  V2a  +  b. 

10.  3  V2a  +  2  V2a  =  5  V2a. 

11.  a/5c  X  9c2  —  VSc  X  16c2  +  \/5c  x  a^  =  (a  —  c)  VSc. 


12.  2a/15  x  i  +  Vl5  X  4  -  Vis  +  Vl5  x  ^  =  fl  Vl5. 


MULTIPLICATION    OF    RADICALS. 


L.  -ea^y'^ 


+  Z<2)3  6a2  (a2  +  ^-2) 


c^  Vcd 

2.  Gab  Vl6a*  x"2c  =  12a2^,  V2c. 


3     /a2  —  Z>2  c2 


3.  6a\/-^ 7-  X  -  =  6a  V(«  +  b)  c. 

\    a  —  b       c 

4.  3a W*  X  5J  X   v'S^  =  15a5  '-v^Sfi^^. 

5.  ^^  X  ^^3^  X  ^-^OF  X  '^W  =  W 


6.  2  VlS"^  X  3  VIOO  =  6  V337500. 

7.  4Vix2^^^  =  8^^^. 

8.  '\/6i  X  'v^  X  'v/125  =  '-^048000. 

9.  WF  X  '^W  X  *^(6F  =  *'^. 

10.  By  the  rule  for  mnltiplication  of  polynomials, 

^  +  8a/J  +  5  Vl  +  ¥  =  ¥  +  Ya/42. 


226-227.]  TKANSFORMATION     OF    RADICALS.  189 

11. 


r9n^  dm  a^  +  h^         Va^  +  bWa^  +  i^ 


a^  +  b^         Va^  +  b^'         '^^^  ^^ 


.-.    Ans.  =  V«M-^' 
12.  The  product  equals 

(\^xy  —  [Vyf  =  x  —y-    ^ns. 

DIVISION  OF  RADICALS. 
2.  2\/3  X  V4  =  2  v^739  x  '^25G  =  2  v^729  x  256, 

2  ^^729  X  256  -  i  ^^8^^  =  4  V^^^=4^^288.  Ans. 

4.     — =r =:  multiplying  both  terms  by  t/a  —  i/b,  we  have 

—        /~ ' 
•  ya  —yb 

again  multiplying  both  terms  by   -y/a  +  ^/b,  we  finally  obtain 


•  a  —  b 


3 


multiplying  both  terms  by  */a  +  l/6^  we  have 

-y/a  4- 2  *ya^  +  -/^     ''^"d      ^'a  —  y/h\ 

multiplying  both  by   -y/a  +  -y/67  <in<3  we  have 

a  +  i  +  2  v^  +  2  X/ZVy  +  2  y"^ 

-r- •     Ant. 

a  —  0 


190 


KEY    TO    DAVIES      BOURDON. 


[227-236. 


6.  y  +  ^V42  =  ^(86  +  13A/42); 

^  ^       ^  2V21  +  6a/2 

Multiplying  both  terms  of  (1)  by  2  V2l  —  G  V2, 

172  \/2l  +  26  V21  X  42  —  516  V2  —  78  a/84 

84  —  72  •  •  ^^ 

26  V21  X  42  =  546  a/2;  -  78  a/84  =— 156  a/21; 


12 


12 


TRANSFORMATIONS    AND    REDUCTIONS. 
Art.  158,  vp.  228-229. 

2.     (3'v/2^)5  =  35^^1(2^5  ^  243^pa)3x(2a)2' 

=  243x2rt\/4^2 

=  4:86a  V  4:0.^-    Ans. 


3. 

4. 
5. 


/" 


(A/a2  —  2a5  -  c)2  =  y  a/(«2  —  2a J  —  c)2 
=  a/«^  —  2«6  —  c.    Ans. 


1. 


^rf.  J59,  p.  229. 

a/8  =  2;    |,/^  =  >v^27  =  |/\/27  =  ^. 

.-.     2\^.     Ans. 
^/^  =  ^ ;     |/>^256  =  v^256  =  ^^ a/256  =  v'i. 


229-236.1       TRANSEOEMATIOXS   AND    REDUCTIONS.  191 

-■  I 


3.     V32aio  =  2a2 ; 


IMAGINARY    QUANTITIES. 
Art.  164,  p.  236. 


a  +  bV—  I 
c  +  dV~^ 


ac  +  5c\/-  1  +  adV-  1  +  M  (V-  l)^ 


=  ac  —  bd  -\-  (be  +  «^)  V—  1.     Ans. 
(2.) 


ffc  —  JcV^  1  —  adV—  1  +  bd{V—  1)^ 

=  ac  —  5c?  —  (a(^  +  Z'c)V— 1.     -4ws. 
(3.) 

=  a2  -  52  +  2a5  V^^ 


(4.) 
flS  +    a2  +    a  +1 


1  +3 -t-_3 +1 

aS  ^.  ^a^V~-^  —  dab^  —  b^  V^^ 

=  a^  —  dal^  +  (3a2J  —  b^)  V^^.    Ans. 


192  KEY    TO    DAYIES'    BOUKDOJST.  [236. 

(5.) 
—  «^  Afis. 


(6.) 
«— 6'y/^ a^-h^J—\  __       a  +  a       _     2a 


{a-bA/-l){a  +  b^-l)      {a-b^-i){a  +  b^/~^)      a'-b'i-l)      a^  +  W 

Ann. 

FRACTIONAL  EXPONENTS. 

1.  a^J~2c-i  multiplied  by  a^^c^ 

hence,  a^j'st-^  x  a^J^c's^  =  ar^b^c~^. 

2.  Sa-^^    multiplied  by    2arhhl 

4_14        21_7  _ 

-3--=-y;      3  +  2  =  ^;      0  +  2  =  2; 

hence,  3a-2jl  x  2a~hic^  =  6a~''^bi(^. 

3.  Qarib^c-""     multiplied  by    baib-^c"^. 
1       1_  _1 
2"^  3~  ~6 

hence,  ea'-^J^-""  x  bah-^C'  =  30a"~6^-'c°-'". 

4.  (o«^)    multiplied  by    ^a^. 

2  2_4        1  2 

3  ^  S^O'      3  ^  3^ 

hence,  (i^  /  ~9 


—  -  +  -=  —  -;    4  —  5=  —  1;     —  m  +  7i  =  n  —  ni; 


239.]  TRANSFORMATION     OF    RADICALS.  UK] 


''      (^"^)=(^^) 


1     1  1    1. 

o  o 

1       1       1_J_,      1       1       1_3 


hence,  ^^cfij=— ai. 

6.     a3     divided  by    oT^. 

2_/_3\_2       3_17. 
'3       \      4/3"^4""l2' 


2 


.i-M 


hence,  as  -^  a*  =  ^ 

7.    a^    divided  by    «*. 

3       4 


4       5  120' 

hence,  ai  -^  a^  =  oT^. 

8.  a6  X  5t    divided  by    cr^s^*^ 

2_/_l\_2       1_^.      3_7_       1 

hence,  a^bi -^  a~Hi  =  aTob-^. 

9.  32ah^ci    divided  by    Sa^Z^^c't 

32-^8  =  4;     1-1  =  1;     6-5  =  1;      ^-|)  =  4;     , 

hence,  33ate8  -j-  8a6^»5c~t  =  4a'^ic* 

10.  64a#c-t    divided  by    32a-9rt(7-t 

64^3.  =  2;9-(-9)  =  18;|-(-|)=5;-^-(-'^)=0 

hence,  64a9jlc~l  ^  32a'^b~ic~i  =  4:aW^. 

11.  Val  X  Vz^i  X  V^. 

hence,  Vaf  x  "v  Ja  x  '^  ci  =  a^b^c^ 

12.  Reduce     ; — ^  to  its  simplest  terms. 

i\/2 


19-i  KEY  TO  DAVIi:s'  BOURDON.  [239. 

CaTicelling  the  -y-^,  and  writing  the  quotient  of  2  -f-  J,    we  hav<» 

43^/37    Ans. 

13.  Reduce        J  — ^i—  v      to  its  simplest  terms. 

(  2V2"(3)^  ) 

Raising  both  terms  to  the  4th  power,  we  have 

aV_2^*  _  li)_*_(3)^V3  _        \/3       _  J_3  y^ 
(2)*  2  (3)'-^       (2)3  (.3)2  (2)^  X  3  ~  384  ^ 

l\    (J)3  _^  y/gl   )  '^ 

14.  Reduce     /  ^  -^^ ^— |  J.-      to  its  simplest  terms. 

V     2V2:(f)^ 


Since  the  square  root  of  the  square  root  is  equal  to  the  4th  root, 
we  need  only  operate  on  the  terms  of  the  fraction  : 

2V2"x  (3)^      V-^i  X  ^-/3  V6 

Multiplying  both  terms  of  the  fraction  by  the  y^  we  have 

\  a  V6  +  v?-)=  \  (1^/^ + ^m 


6 


:  hence, 


V4(iV6+  v/21.) 


15.  Multiply     a^+  a^-^    +  A    -\-  ah  ^  aV  +  6* 


i>y 


a    —  0    ; 

■h^ 

cfi  —  }p-.     Ans. 


240.] 


TRANSFORMATION,  -QF    RADICALS. 


195 


16.    Divide    a^  —  a?h  ^  —  a^b  +  b^    \\  a^  —  b' 


17. 


18. 


19. 


20. 


«t- 

a=^r^                 «'  -  ^ 

—  ah  +  h^ 

-  ah  +  b^. 

>x^  —  fa* 

a;^  -  \a^ 

xi  —  ^h^ 

x^  4-  ^a^a;"^  4-  ^a;~*  4-  &c. 

JL  1.                    1 

^a^x^  —  lax  * 

+  iax~^  - 1«* 

&c.,        &c. 

a;-i+a;~%  2+  7/"^ 

2/    +a;V      +^ 

yx-^  +  a:~ 2^2  _|_  i 

x~hj^  +  1  +  a;^2/~^ 

4-  1  4-  a;^^/"  8  +  a;«r^ 

2/a;-i  +  2a;"2^y^+  3  +^x^y~^  +  a;^^-! 

/a^      ^»"^\2     a^     ^ah^       b^ 
Is  ~3  /-4  ""^     6      '9 

=  !«!_  1,1,1+ iji 

a;-!  +  x~^y~^  +  JT^ 

a;-!  +  a;~^i/~ «  +  ^1 

a;-2  +  x~^y~^  4-  x-^y-^ 

—  a;"2|^~2'  —  x-^y-'^  —  x~^ij~^ 

+  a:-i2/-i  +  x-^y'i  4-  ^^ 

ar2 


+  a;-iy-i 


+  2/ 


-2 


196 


KEY   TO   DA  vies'    BOURDON. 


[240. 


21.  {x  ^  -y   3)2  =  x-^  -  2x  ^y~^  +  y-\ 

22.  x^  —  4a;^  +  3 


X 


■X'^ 


X 


.1 


4a;^  +  2x 


+  4:X  —  x^  —  2x^ 


x^  —  Axi  +  6a;  —  a;2  —  2xk 
23.  ^x-^-^x^  +  iJ-^x^-^x-^  +  ^x^ 

-\x^—hx^+^x^ 


X 


■4 3^i 


^^x^-^x^+^x^ 


\x 


.1^ 


\x 


•-^s 


—XtJ^ 


24.     a?-{-  xy  +  y'^ 

a;  +  x^y^  +  ?/ 
a;^4-  x^y  +  a^T/^ 

+  2^2^  +  xy^ 


-^x^-^^x^ 

—hx^  +  \x^. 
0 


+  x^y^  +  a;^t/^  +  a; V 


+  2/^ 


a:3+  2a:2?^  +  2xy'^  +x^y^  +  a;%^  +  x^y^  +  ^3. 

25.     a;^  +  2a;^  +  3a;^  +  23:^  +  1 
a;i  _  2a;^  +  1 

a;    +  2a:^  +  3a:^  +  2a;^  +    x^ 

—  2x^  —  4a;^  —  6x^  —  Ax^  —  2x^ 
+    a;^  +  2x^  +  3a;^  +  2a;^  +  1 


X 


2x^ 


+  1. 


240-248-249.]  arithmetical  progbessiojs".  197 

26.  x^  -f  2x^  +  dx^  +  2a;^  +  1 

xi  —  2.g^  +  1 

X    +  2x^  +  dx^  +  2x^  +    x^ 

—  2x^  —  4:X^  —  6x^  —  4:x^  —  2x^ 

x^  +  2x^  +  Sx^  +  2x^  +  1 

z  —2x^  +1 

ARITHMETICAL  PROGRESSIOK 

2.  a  =  2,     d-l,     «=:100;     .  • .    ^  =  a  +  (n  -  1)  <;  =  695. 

3.  a-\,     c?=2,     n  1=100;     .- .    I  =  a -\- {n  -  \)d  =z\^^. 
Hence,  ^  =  ^w  (a  +  ;)  =  10000. 

4.  ;  =  70,     c?  =  3,     n  =  21, 

a  =  l-{n-\)d=\0',     S  =  :^n[2l  ~{n-  l)i]  =  840. 

5.  a  =  10,     d=-\,     n  =  21     .  ■ .  I  =  a  +  {n  -  l)d  =  ^; 
whence,  S  =  ^n  [21  —  (?i  —  l)d]  =  140. 

6.  d  =  6,     1=  185,     ^  =  2945 


2l  +  d  ±  ^{21  +  rf)^  -  St^^      376  ±  4 
*"-  2d  -        12 

taking  the  lower  sign,     n  =  31,         a  =:  Z  —  (n  —  \)d  =  5. 

7.  a  =  2,     Z  =  5,     «  =  11;     .  • .  (f  =  in^  ==  £  =  0.  3. 

7i —   1  10 

8.  a=l,     c?=l,     n  =.  n     I  =  n  ', 

.'.  .   S  =  in[2a4-(n-l)d]  =  ^^  "^  "^  ^ 

9.  a=l,     (Z  =  2,     n  =  /i;     . ' .    5  =  ^n  [2a  +  (n  —  l)d]  =  n» 
10.     a  =  4,     (/ =  4,     n  =  100; 

.    .     ;S  =  ^«  [2a  +  (n  -  1)(£]  =  20200 
20200yr/i-,  =  llmi.  840ycif. 


198  KEY  TO  DAVIES'  BOUKDON.     [253-54-56   58. 

GEOMETRICAL   PROGRESSION. 
8.     a  =  2,     r  =  3,     /  =  39366 

S  =  ^1:Z^  =  59048. 
r  —  1 

4.  a  =  1,     r  =  2,     n  =  12; 

.'.     S  = z=  4095  ;     I  -  ar''-^  =  2048. 

r  —  1 

5.  a  =  1,     r  —  2,     n  =  12  ; 

.  • .     *S^  = =  4095  ;  4  >95.9.  =  £204  I5«. 

r  —  1 

6.  <i=  1,     r  =  3,     n  =  10; 

„       ar*—a       59048  ^^.  ^, 

.  • .     S= =  — -—  =    295.24. 

r  —  1  2 

I  _  ay.«-i  _     196.83. 

INSERTING    GEOMETRICAL    MEANS. 

2.     a  =  2,     i  =  486,     m  =  4  ;         .  • .     r  =  ^/243  =  3  : 
hence,  the  progression,         2  :  6  :  18  :  54  :  162  :  486. 

SUMMATION   OF  SERIES. 
4.  y  =  4,     ^  z=  4,     n  =  1,     5,     9,     13,     &c. 

4       4       4        4         4 

1st  auxiliary  series,     -+-  +  --(-_-f_-f  ... 

4       4        4         4  4 


5   '   9       13   '    17    ■  "^  4»  — 3' 


287-311.]  EXPONENTIM.    Et^UATrONS.      .  199 

PILING   BALLa 

1.  „  =  i6;      ...S=2i^i^±i)  =  680. 

8.    „=Ui      ...^^''("+')(^"+')^ioi5; 

n'  =  5,     ^'  =  55  ;       .'.   S-  S'  =  960. 

3.  «  =  30,     «  =  30 ;      .  • .  S  =  ^(2±21(i±|f±5!L)=  23405. 

4.  m  =  26,     n  =  20  ;        .  • .   -S  =  8330  ;     m  =  26,     n'  =  S  ; 

6"  =  1140;      .-.   S  -  S'  --^  7190 

5.  n  =  20;  .-.  /S  =  1540;    n'  =  9;  .-.  S'  =  165; 

.-.  ^— ;S^'  =  1375. 

6.  M  =  15;  .-.  5=  1240;    w' =  5;  .-.  ;S"  =  55; 

.'.  S-S'  =  1185. 

7.  m  =  52,  n  =  40;  .-.S  =  64780;  m'  =  52;  w'  =  17; 

.-.  ;S' =  9741 ;.-.  >S  — /S' =  55039. 

EXPONENTIAL  EQUATIONS. 
These  equations  may  be  solved  as  the  preceding  ones  have  been, 
but  it  will  be  better  to  make  use  of  the  table  of  logarithms  (P.  291), 

8*  =  32 ;    hence  we  have, 

log  32      5 

a;  log  a;  =  log  32;    or,    ^  =  "i^g- =  3 

2.  3'  =  15  ; 

taking  the  logarithms  of  both  members, 
1      o       1      ,r  log  15       1.176091       „    _ 


200  •  KEY  TO  DAVIKS'  BOUKDON.  [311-331. 

3.  10' =  3; 

taking  the  logarithms  of  both  members, 

X  log  10  =  log  3,     or    a;  =  log  3  =  0 .  477121. 

taking  the  logarithms  of  both  members, 

a;  log  5  =  log  f  ^  log  2  -  log  3  ; 

log2-l(.g3  _  -  .170091 
•'•     ""^        logs        -       .(398970  '-"  ~  ^•^* 


THEORY   OF  EQUATIONS. 

2.  Two  roots  of  the  equation, 

X*  —  12x3  +  48a:2  _  QSx  +  15  =  0, 

are  3  and  5  :  what  does  the  equation  become  when  freed  of  them  ? 

x*  —  12^^  +  48x2  _  68x  +  15  \x  -  S 

x^  —  Vx^  +  21x  —  5 

x2  -    9x2  +  21x   —  5  |x   -  5 

x2  —  4x  +  1  =0.     Ans. 

3.  A  root  of  the  equation, 

ar3  —  6x2  4-  liar  —  6  =  0, 

is  1  :  what  is  the  reduced  equation  1 

x^  —  6x2  4-  iia;  _  6  |x  —  1 

x2  —  5x  +  6  =0.     Ans 

4.  Two  roots  of  the  equation, 

4x*  —  14x3  _  5^2  -f  3I2;  +  6  =  0, 
are  2  and  3  ;  find  the  reduced  equation. 


336-337-340.]     greatest  common  divisor.  201 

4x4  _  14^,3  _  5a.2  +  31^  ^  (5  \x  -2 

4a:3  _  6^2  _  17^.  _  3  |  ^^  _  3 

4.r2  +  6x  4-  1  =  0.     Ans 

FORMATION   OF  EQUATIONS 

2.  What  is  the  equation  whose  roots  are  1,  2,  and  —  3  1 

(x  -1)  {x  -2)  {x  +  3)  =  x^  -7x-\-G=  0.     ^ns. 

3.  What  is  the  equation  whose  roots  are  3,    —  4,    2  +  V^'   ^^^^ 
2-  "v/al 

(a;  -  3)  (.r  +  4)  (a;  -  2  -  -/3)   (x  -  2  4-  V^) 

_  a.-*  -  3x3  _  15^.2  ^  49j.  _  12  =  0.     ^«jf. 

4.  What  is  the   equation   whose   roots  are    3  +  y^,  3  —  -y/5^ 
and   -  6  1 

(.r  -  3  —  v^)  (x  —  3  +  -v/S)  (x  +  6)  =  x3  —  32x  -f  24  =  0.    Ans. 

5.  What   is   the  equation  whose  roots   are    1,  —  2,  3,   —  4,  5. 
and  -  6  ? 

{z  -  1)  (x  +  2)  (x  -  3)  (x  +  4)   (x  -  5)  (x  +  6) 
=  x«  +  3x5  _  41^.4  _  87a;3  +  400x2  +  444x  —  720  =  0.     Ans, 


Q.  What  is  the   equation   whose    roots    are  ....  2  -f-  y-—  It 
2  -  -/~^^'  and   -  3? 

(x  -  2  -  V^=T)   (-»•  -  2  +/^^  (x  +  3) 
=  x3  —  x2  —  7x  4-  1«^  =  0.     Ans. 

(a;_l_V-l)  (.i:_l  +  V-r)   =   {x-iy-{V-ly 
—  ^  _2x  -\-  2; 

.-.{x—V-l)  {x+V-l){x-l  —  V-l){x-l  +  V^^){x-l) 

=  (a;2  +  l)  (x^—2x  +  2)  (x—l)  =  0. 

a;5_3a;4_^52.3_5^.2  +  4a;— 2  =  0. 


202  KEY    TO    DAVIES'    BOCEDOlf.  [340-341. 

TRANSFORMATION    OF    EQUATIONS. 

1.  Transform  the  equation, 

a^  ^  dx*  —  4:X^  -]-  x^  —  X  +  4:  =  0, 

into  one  whose  roots  are  equal  to  those  of  the  given  equation 
with  their  signs  changed. 

Making  x  =  —  y  and  dividing  by  —  1, 

^5  —  3?/*  —  4?/3  —  2/2  —  2/  —  4  =  0. 

1.  Transform  the  equation, 

a^  +  3a:2  +  a;  +  ^  =  0, 

into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    x^- ,    substituting,  and  reducing, 

2/3  +  3^/2  4-  9?/  +  3  =  0. 

2.  Transform  the  equation, 

a;4  ^_  2.3  _|_  3,^  ^  2  =  0, 
into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    x=z  -,    substituting,  and  reducing, 

.      3,1         1       ^ 

3.  Transform  the  equation, 

7         7 
^+2^-3  =  °' 

into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    a;=-,     substituting,  and  reducing, 

2  ^  A 

y^-y-~r,  =  0. 


347-348.]  TKANSFOKMATION    OF    EQUATIONS.  203 

TRANSFORMATION    OP    EQUATIONS    BY    SYNTHETICAL 

DIVISION. 

4,  Find  the  equation  whose  roots  shall   be   less  by  3  than 
the  roots  of  the  equation 

a^  _  3a;3  _  i5a;2  +  49^;  _  12  =  0, 

1  —  3  —  15  +  49  —  12  |]3 
+  3  +    0  —  45  +  12 


+  0  -  15  + 

4,+ 

0 

+  3+  9- 

18 

+  3-  6,- 

14 

+  3  +  18 

+  6  +  12 

+  3 

1,+  9    .-.    2/4  +  92/3  +  l2?/2  —  Uyz=0 

5.  Find  the  equation  whose  roots  shall  be  less  by  10  than 
the  roots  of  the  equation 

a;4  +  2a^  +  3a;2  +  4a;  —  12340  =  0. 

1+2+      3+        4  —  12340  pO 
+  10  +  120  +  1230  +  12340 


+  12  +  123  +  1234,+ 0 

+  10  +  220  +  3430 

+  22  +  343,+ 4664 

+  10  +  320 

+  32,+  663 

+  10 

1,  +  42    .  • .    y^  +  422/3  +  663^2  +  46642/  =  0. 


204  KEY    TO    DAVIES'    BOUKDON.  [348-349. 

6.  Find  the  equation  whose  roots  shall  be  less  by  3  than 
the  roots  of  the  equation 

x^  -\-2x^  —  6a;2  —  10a;  =  0. 

1+0+2-6-10  +  0^ 
+  2+    4  +  12+12  +  4 


+  2+  6+  6+  2,+ 4 

+  2  +  8  +  28  +  C8 

+  4  +  14  +  34,+  70 

+  2  +  12  +  52 

+  6  +  26,+  86 

+  2  +  16 

+  8,+ 42 

+  2 

1,+  10;  .-.  2/5  +  10y4  +  42j/3  +  86?/2  +  70?/  +  4=0. 

DISAPPEARANCE  OF  SECOND  TERM. 

2.  Transform  the  equation 

a^  —  lOx*  +  7a:2  +  4a;  —  9  =  0 

into  an  equation  in  which  the  second  term  shall  be  wanting. 
Here  we  make  the  roots  of  the   resulting  equation  greater 

P 
than  those  of  the  given  equation  by    — -r,    or    —2    (Bour- 

0 

don,  Art.  266) ;  that  is,  we  make  them  less  than  those  of  the 
given  equation  by     +  2.    Hence, 

OPEEATION. 

1_10+    7+    0+    4-    9L2_ 
+    2  —  16-18-38-64 


349.]  TRANSFORMATION    OF    EQUATIONS. 

1st  quotient, 


205 


2d  quotieut, 
3d  quotient, 
4th  quotient, 


1_8—    9—    18—    32,—  73      1st  rem. 

+  2  —  12  —   42  —  120 
1  _  6  —  21  —    60,—  152      2d  rem. 

+  2  —    8  —    58 
1  _  4  _  29,—  118      3d  rem. 

+  2—4 
1  _  2,—  33     4tii  rem. 

+  2 


5th  quotient,       1,+  0     5th  rem. 

Hence,  the  transformed  equation  is 

y5  _  33^3  _  1182/2  _  152?/  _  73  =  0. 

3.  Transfonn  the  equation 

jr3  _  6a;2  +  7a;  —  10  =  0 

into  one  whose  second  term  shall  be  wanting. 

Here  we  make  the  roots  of  the   resulting  equation  greater 

P 

than  those  of  the  given  equation  by     —  -^  =  —  2 ;     that  is, 

we  make  them  less  than  those  of  the  given  equation  by   +  2. 


1st  quotient, 
2d  quotient, 
3d  quotient, 


OPEEATION, 

1-6  +  7-10^ 
+2—8—    2 

1-4  —  1,-12    1st  rem. 

+  2  —  4 

1  —  2,—  5    2d  rem. 

+  2 


1,     0     3d  rem. 


206  KEY    TO    DAVIES'    BOURDON.  [349. 

Hence,  the  transformed  equation  is 

y3—5y—l2=  0. 

4.  Transform  the  equation, 

into  one  whose  second  term  shall  be  wanting. 
Here  we  make  the  roots  of  the   resulting  equation  greater 

P 

than  those  of  the  given  equation  by     +  —  =  3 ;     that  is,  we 

o 

make   them   greater  than   those  of  the  given  equation  by  3. 
Therefore,  the  synthetical  divisor  is     —  3. 

OPERATION. 

14-9—    1  +    4[— 3 
—  3  —  18  +  57 
1st  quotient,        1  +  6  —  19,+  61     1st  rem. 
1  —  3—    9 


2d  quotient,         1  +  3,-28    2d  rem. 
—  3 


3d  quotient,         1,  +  0    3d  rem. 
Hence,  the  transformed  equation  is 
2/3  —  28?/  +  61  =  0. 

5.  Transform  the  equation 

a;*  —  8a^  +  7a:2  +  3a;  +  4  =r  0 

into  one  whose  second  term  shall  be  wanting. 

Here  we  make  the  roots  of  the  resulting  equation  greater 

P 

than  those  of  the   given  equation  by     —  —  =  —  2 ;    that  is, 

we  make  them  less  than  those  of  the  given  equation  by  +  2 ; 
hence,  the  synthetical  divisor  is     +  2. 


t9-359.J 

EQUAL    ROOTS. 

OPERATION. 

1-8+    7+    3+    4   3 
+  2  —  12-10  —  14 

1st  quotient, 

1  _  6  —    5  -    7,-10     1st  rem. 
+  2—    8  —  26 

2d  quotient, 

1  _  4  —  13,—  33    2d  rem. 

+  2—4 

3d  quotient. 

1  —  2,—  17    3d  rem. 

+  2 

4th  quotient, 

1,+  0    4th  rem. 

207 


yi  _  17^2  _  33^  _  10  =  0. 


EQUAL    ROOTS. 


4.  What  are  the  factors  of  the  first  member  of 

a;7  +  5a;6  +  Qr^  —  Qx^  —  15a;3  _  32:2  _|_  ga;  +  4  =  0  ? 
The  first  derived  polynomial  is 

"iofi  +  30a:5  +  ^Qx^  —  Ua?  —  45a;2  _  62;  +  8, 
and  the  common  divisor  is 

z!^  +  da?  +  x^  —  Zx  —  % 
The  equation 

a4  +  32;3  +  a^  _  3a;  _  2  _  0,    .     .    .    . 


can  be  factored  by  the  method  of  equal  roots, 
derived  polynomial  is 

4a:3  +  9a;2  +  22;  -  3, 


•    .     (1) 

The  first 


and  the  common  divisor  is  a*  +  1  ;  that  is,  a;  +  1  is  twice  a 
factor  in  the  first  member  of  (1).  The  remaining  factor,  found 
by  division,  is    a^  +  a;  —  2,  or   (a;  —  1)    (a;  +  2).      As  each 


208  KEY    TO    DAYIES'    BOURDO^-.  [359. 

factor  is  contained  once  more  in  the  given  equation  than  it 
is  in  (1),  we  have, 

{x  +  1)3  {x  —  1)2  {x  +  2)2  =  0. 

5.  What  are  the  equal  factors  of  the  equation 

x!  —  7o^  +  10^:5  +  22xi  —  43x^  —  Soz^  +  482;  +  36  =  0  ? 

The  first  derived  polynomial  is 

7x^  —  42:^5  ^  50:^4  ^  ^^^  _  i^Qx^  —  70a;  +  48, 

and  the  common  divisor  between  it  and  the  first  member  of  the 

given  equation,  is 

xi  —  3a^  —  3x^  +  7x  +  5. 
The  equation 

x^  —  Sa^  +  Sx^  +  ^x  -{-  6  =  0, 

cannot  be  solved  directly,  but  by  applying  to  it  the  method  of 
equal  roots,  that  is,  by  seeking  for  a  common  divisor  between 
the  first  member  and  its  derived  polynomial, 

4a^  —  9a;2  —  6x  +  7, 

we  find  such  divisor  to  be  x  -\-  1;    hence,   a;  +  1   is  tivice  a 
factor  of  the  first  derived  polynomial,  -and  three  tiines  a  factor 
of  the  first  member  of  the  given  equation  (Art.  271). 
Dividing, 

x^  —  3u^  —  dx^  -^  llx  -\-  6  =  0,  by  {x  +  1)2  =  a;2  +  2a;  +  1, 
we  have,  a;2  —  5a;  +  6,  > 

which  being  placed  equal  to  0,  gives  the  two  roots 

a;  =  2    and    x  =  3, 
and  the  two  factors,    x  —  2     and    x  —  3. 

Therefore,  {x  —  2)  and  (x  —  3),  each  enters  twice  as  a  factor  of 
the  given  equation ;  hence,  the  factors  are 

{x  -  2)2  (a;  -  3)2  {x  +  1)3.     Ans. 


359-384.]  COMMENSURABLE    ROOTS.  209 

6.  What  are  the  equal  factors  of  the  equation 

x!  —  3afi  ^9x^  —  IQx^  +  27a^  —  33a;2  +  27a;  —  9  =  0  ? 

The  first  derived  polyuomial  is 

7x^  —  182^  +  45a;*  -  76a;3  ^  ^i^  _  qq^  +  27, 

and  the  common  divisor  between  it  and  the  first  member  of  the 
given  equation,  is 

2^  —  2a^  +  4:X^  —  6x  -{-  3. 
The  equation 

xi  —  2a^  +  Ax-  —  6x  +  3  =  0 

cannot  be  solved  directly ;  but  by  applying  to  it  the  method  of 
equal  roots,  as  in  the  last  example,  we  find  the  derived  poly- 
nomial to  be 

4:c3  _  6a;2  -f-  8a;  —  6, 

and  the  common  divisor  to  be  x  —  1 ;  hence,  (.t  —  1)  enters 
twice  as  a  factor  into  the  derived  polynomial,  and  three  times 
as  a  factor  into  the  first  member  of  the  given  equation. 

Dividing 

cc*  —  2a;  +  4a^  —  6a;  +  3     by     (a;  —  1)2  =  a;2  —  2a;  +  ], 

we  have  for  a  quotient  a;^  -f  3  ;   hence,    {x^  +  3)    enters  twice 

as  a  factor  into  the  first  member  of  the  given  equation  ; 
hence  the  factors  are 

{x  -  1)3  (a?  +  3)2. 

COMMENStTRABLE    ROOTS. 

2.  What  are  the  entire  roots  of  the  equation 

a:4  _  5a;3  ^  25r  —  21  =  0  ? 

The  divisors  of  the  last  term  are  +1,  —  1,  +3,  —  3,  4-7, 
—  7,  +  21,  and  —  21 ;   i^  =  22  ;    —  L"  =  —  4. 


210  KEY  TO  DAVIES'  BOURDON.  [384- 

+  21,     +7,  +3,  +1,  -1,  -3,     - 

-    1,      -  3,  -  7,  -  21,  4-  21,  +  7,     -h 

4-  24,     +  22,  +  18,  +    4,  -f  46,  +  32,  + 

+    3,  +    6,  +    4,  -46,  - 

+    2,   -f     4,     +  46, 

-3,-1,     +  41, 

-    1,  -    1;    -41, 
therefore,    +  3  and    +  1   are  the  two  entire  roots.     Dividing  th« 
first  member  of  the  equation  by  the  product  of  the  factors 

(x  —  3)  {x  —I)  =  x^  -  4x  +  3, 
we  have  x'^  —  x  —  y  =  0. 

Note. — In  the  4th  line  we  add  the  co-efRcient  of  x',  which  is  0,  and  then 
divide  by  the  divisors,  and  thus  obtain  the  6th  line. 

3.   What  are  the  entire  roots  of  the  equation 

I5x^  -  lOx*  +  6x3  +  15x2  _  19a;  -I-  6  =  0  ? 

■f    6,     +3,     +2,     +1,     -    1,     -    2,     -    3,     -    6, 

-f-    1,     +    2,     +    3,     +    6,     -    6,     -    3,     -    2,     -    1, 

-18,     -17,     -16,     -13.     -25,     -22,     -21,     -20, 

-    3,  -    8,     -  13,      ^25,     4-  11,     +    7, 

-f-  12,  +    7,     +    2,     +  40,     +  26,     +  22, 

-♦-2,  +    2,     -  40,     -  13, 

«-    8,  +8,     -34,     -    7, 

+    8,     +34, 

-11,     +15, 

-11,     -15, 

+  15: 

hence,  there  is  but  one  entire  root,  which  is   —  1 . 


384-392.]  NUMBER    OF    REAL     ROOTS.  211 

4.  What  are  the  entire  roots  of  the  equation 

9a^  +  30a;5  _(_  22a;4  +  lOar^  _,_  i7a;2  _  20a;  +  4  =  0? 

This  is  worked  like  the  preceding  example,  giving  the  en- 
tire root,  —  2.  Then  dividing  the  equation  by  x  +  2,  we 
find  a  new  one,  which  has  a  root,     —  2. 

NUMBER  OF  REAL  ROOTS. 

3.  What  is  the  number  of  real  roots  of  the  equation 

a;3_5^^8a;  — 1  =0? 

By  finding  the  expressions  which  indicate,  by  their  change  of 

sign,  the  existence  of  real  roots  (Art.  293  and  Example  1),  we 

have 

X=    x^—    5a;2+8a;  — 1 

Xr=Sx^  —  10x  +  8 

Xj=  2x  —  31 

X3=  —  2295 

x=  —  cc    gives 1 2  variations, 

a;  =  +  00    gives     +  +  H 1  variation ; 

hence,  there  is  one  real  and  two  imaginary  roots  (Art.  293). 

For    X  =  0,    we  have 1 2  variations, 

for      a;  =  1,         "  +  -| 1  variation ; 

hence,  the  real  root  lies  between  0  and  +1. 

4.  Find  the  number,  places,  and  limits  of  the  real  roots  of 

x*—8a^-\-  Ux^  +  la;  _  8  =  0. 
For  solution,  see  Example  3,  page  141  of  Key. 

5.  Find  the  number,  places,  and  limits  of  the  real  roots  of 

the  equation 

a;3  —  23a;  —  24  =  0. 


;^13  KEY    TO    DAVIES'    BOURDON.  [393-^96. 

X  =  rc3  _  23^  _  24 
Xi=3a;2-23 
X,=  23a;  +  36 
X3=  8279. 

For    x=  —cc,    we  have, \ \-,    3  variations. 

For    a;=  +  00,    we  have,     +  +  +  +?    no  variations. 
Hence,  there  are  3  real  roots,  which  are  easily  placed. 

6.  In  the  sixth  example,  we  have, 

X  =  a^  +  jx^  —  2x  —  6 
X,=:3a;?+3a;  — 2 
X2=  11a;  +  28 
X3=  —  1186. 

Hence,  there  is  but  one   real   root,  and   that  lies  between 
the  limits  1  and  2. 


CUBIC  EQUATIONS. 

1.  What  are  the  roots  of  the  equation 

x^-Qx^+Zx  =  lS     •     .     .     (1)1 
Transforming  so  as  to  make  the  second  term  disappear, 
a;3_9a:_28  =  0     •     ■     •     (2) 
p=-9         q  =  -2B; 
substituting  in  Cardan's  formula,  and  reducing, 

X  =  4. 
But  the  roots  of  the  given  equation  are  greater  than  those  of  equa- 
tion (2)  by  2 ;  hence  ar  =  (>, 


396-405.]  CUBIC  kquations.  213 

Transposing    18   in  equation  (1),  and  dividing   both  numbers  by 

«  —  6,  we  find 

a:2  4-3-0         .-.  x=:  ±  ^-  3; 


hence  the  three  roots  are         6,     -y/  —  3,     —  ^  —  3. 

2.  What  are  the  roots  of  the  equation 

a;3  _  9j;2  4.  282;  =  30     ....     (1)? 
Transforming,  we  find 

a;3 -H  a;  =  0  (2)      .-.     a;  =  0     and     a;  =  ±  y'—  1- 
But  the  roots  of  (1)  are  greater  than  those  of  (2)  by  3; 

hence  the  roots  of  (1)  are,     3,  3  +  -y/  —\     and     3  —  |/—  1. 

3.  a:3  —  72;  +  14  =  20 (1) 

Transforming  (Bourdon,  Art.  266),  we  have, 

„      7         344      c^(^^       X.  ^  344 

3'^~T"~^^'       ^°^^'    ^~~3'     ^~~W 

g 
Substituting  in  Cardan's  formula,  we  have  - ,  the  real  root. 

o 

7 
But  the  roots  of  (1)  are  greater  by  -    than  those  of  (2) ;  hence, 

o 

in  (1)    a;  =  5. 

Transposing  and  dividing  by    a;  —  5,    we  have, 

a;2  _  2a;  +  4  =  0 ; 
hence,  the  required  roots  are,  5,   1  +  ^/~^^,  and  1  —  V —  3. 

HORNER'S  METHOD  OF  SOLVING  NUMERICAL  EQUATIONS. 
1.  a?  +  x^  +  x  —  100  =  0. 

By  Sturm's  Rule,  we  find 


214 


KEY  TO  DA  VIES    BOUEDON 


[405. 


variations, 
variation  ; 

variations, 
variation ; 


X  ^  x^+x^  +  X   -  100, 
Xi  =       3x2  +  2x  +  1 
Xj  =  —  4x   +  899 
X3  =  -  2409336. 

For         X  =  —  (X),  ^.-f._j  2 

for  xz=  +  cc,  +H ,  1 

hence,  there  is  but  one  real  root. 
i^or         z  =  4,  _  +  4-  _  ^  2 

for  2;  =  5,  4.  +  4.  _  ^  1 

hence,  the  real  root  lies  between  4  and  5. 

2.  2*  -  12x2  +  12x  -  3  =  3 

By  Sturm's  Rule, 

X  =x*    -  12x2+  12x-  3, 

Xi  =  4x3  _  24x  +  12,     or     x^  —  6ar  +  3,     * 

X^  =  2x^  —    Sx+  }, 

X3  =  13x—    9, 

X^  =  20. 

For  a:=z— Qo,  -| 1 ^.,  4  variation  *, 

for  2;  =  +  00 ,  -I-  +  +•  4-  4.  ^  0  variatio*  j 


hence,  there  are  4  real  roots. 


For     X  =  —  4 


for 
for 
for 
for 
for 


X  =  -3 

X  =  0 
X  =  +  1 
X  =  +2 
X  ^  4-  3 


4-  -  +  -  + 

+  -  + 

-  +  +  -  + 

=F   +    + 

—  +  +  + 

+  +  -^  +  + 


4  variations, 
3  variations, 
3  variations, 
1  variation, 
1  variation, 
0  variation ; 


406.] 


HORNER  S    METHOD. 


215 
3,  two  between  0 


hence,  one  of  the  roots  lies  between  —  4  and 
and  1,  and  the  remaining  root  lies  between  2  and  3. 

3  x*  —  8x^  +  14x2  +  42;  _  8  _  0. 

By  Sturm's  Rule, 

X  =x*    -  8a;3    +  14x2  +  43.  _  g, 

JTj  =  4x3  _  24x2  _f.  28x  +  4,     or    X''  —  6x2  4.  7^;  4.  j^ 

Xj  =  5x2  _  17a.  4.  6^ 

X3  =  76x  -  103, 

X4  =  45475. 

For    2;=—  CO,         + \- 1- 

for      a;=  +  QO,         +  +  +  +  + 
hence,  the  equation  has  4  real  roots. 


4  variations, 
0  variation ; 


For 

x=-l             4-  _  +  _  + 

4  variations. 

for 

x=       0             _  +  +  _  + 

3  variations, 

for 

x=  +  l              +  + + 

2  variations, 

for 

x=+2             + +  + 

2  variations. 

for 

x=  +S             ±+  + 

1  variation. 

for 

x=       5             +  +  + 

1  variation, 

for 

x=       6             +  +  +  4-  + 

0  variation  ; 

hence,  one  root  is  between  —  1  and  0,  one  bet 

;ween  0  and  1,  one 

between  2  and  3,  and  one  between  5  and  6. 

4. 

a;5  _  10x3  +  6x  +  1  =  0. 

By  Sturm's  Rule, 

X  =x^      -  10x3+  6x  +  1, 

Xi  =  5x*    -  30x2+  6, 

Xj  =  20x3  _  24x  -  5, 

X3  =  96x2  -  5x    -  24, 

X^  =  43651X  +  10920, 

X^  =  32335636224. 


216 


KEY  TO  DAVIES    BOURDON. 


[40& 


For     a;  =  —  ao  —  +  — 

for      X  =  +  CD  +  +  + 

bence,  the  equation  has  5  real  roots, 


For     X  = 


for 
for 
for 
for 
for 
for 


hence, 


x=  —S 

x=  —  1 
X  =  0 
x  =  +  1 
a;=  +3 
a;=:  -f-4 


-  +  - 

+  +  - 

+  +  - 

+ 

+  +  -I- 


one  root  lies  between 
two  roots  lie  between 
one  roDt  lies  between 
one  rojt  lies  between 


+  -  + 

+  +  + 

+  -  + 

+  -  + 

+  -  + 

—  +  + 
+  +  + 
+  +  + 
+  +  + 

—  4  and 

—  1  and 
0  and 
3  and 


5  variations, 
0  variation  j 

5  variations, 
4  variations, 
4  variations, 
2  variations 
1  variation, 
1  variation ; 
0  variation ; 

-3, 

0, 

+  1, 
4. 


APPENDIX. 


GENERAL  SOLUTluK  OF  TWO  SIMULTANEOUS  EQUATIONS  OF  THE 

FIRST  DEGREK 
1.  Take  the  equations, 

ax  -\-   by  ■:=  c      .     .     .     (1), 

a'x  -\-b'y  =  c'     .     .     .     (2) ; 

multiply  both  members  of  (1)  by  b'  and  of  (2)  by  b,  then  sub- 
tracting and  factoring,  we  find 

{ab'  -  a'b)  x  =  b'c^  be' ; 

b'c  -  be'  .„. 

•••  '  =  W^:i^b  •  '  '  (^)- 

I    11  ac'  —  a'c 

In  like  manner,  y  =  -r^ tt     •     •     •     (4). 

^        ab'  —  a'b  \  / 

By  means  of  formulas  (3)  and  (4)  any  two  simultaneous  equations 
of  the  forms  (1)  and  (2)  may  be  solved. 

Thus,  4a:  +  3y  =  31, 

3a;  +  2?/  =  22  : 

by  comparison  with  (1)  and  (2), 

a  =  4,     6  =  3,     c  =  31,     a'  =  3,     b'  =  2,     c'  =  22 ; 

by  substitution  in  (3)  and  (4), 

62  —  66       ,  88  —  93      ■ 


218  APPENDIX. 

EXAMPLES. 

(     -+^    =2     ) 
1.  Given  J     3      4  l       to  find  x  and  y. 

(  3a;  +  4y  =  25   ) 
B/  comparison  with  (1)  and  (2), 

«  =  i,         *   =  i,         c  =    2, 
a'  =  3,         6'  ^  4,         c'  =  25  ; 
by  substitution  in  (3)  and  (4), 

8    -  6i       ^  8i  -  6       ^ 

'3  4  ^3  4 


(        \\x  —    hy  =  —\\ 
2.  Given  ■<  >•      to  find  x  and  y : 

(-  5a:    +  16y  =  124)  ' 

by  comparison, 

a   =        11,         6   =  —    5,         c  =  —  1, 
a'l^  -    5,         6'=       16,         c'=       124; 

oy  substitution, 

_  -  16    +  620  _  _  1364  —  5    _  ^ 

^  ~        176  -  25    ~    '         ^  ~    176  -  25  ~  '  * 


GENERAL  SOLUTION  OF  THREE  SIMULTANEOUS  EQUATIONS  OF 

THE  FIRST  DEGREK 

2.  Take  the  equations, 

ax    •\-  hy    -\-  cz    z=.  d     •     •     •      (1), 
a'x  +  b'y  -\-  c'z  =  d'    '     •     •     (2), 
a"x  +  b"y  +  c"z  =  d"  -     •     •     (3). 
From  (1)  and  (2)  we  obtain,  by  eliminating  2, 

(</a  -  ca')  X  +  {c'b  -  cb')  y  —  c'd  —  cd'     •     ■     ■     (4). 


ADDITIONAL   EXAMPLES.  219 

In  like  manner,  from  (1)  and  (3), 

{c"a  -  ca")  X  +  (c"b  -  cb")  y  =  c"d  -  cd"         -     •     •     (5j  ; 
combining  (4)  and  (5)  and  eliminating  y,  we  find 


_  (c"b  -  cb")  (c'd  -  cd')  -  (c'b  -  cb')  (c"d  -  cd") 
{c'a   -  ca'  )  {c"b  -cb")  -  {c"a-  ca")  {c'b  -  cb') 


X  = 


(6). 


In  like  manner, 


_  {c'a  -  ca')  {c"d  -  cd")  -  {c"u-  ca")  {c'd  -  cd') 
y  -  [c'a  -  ca')  {c"b  -  cb")  -  {c"a-la")  [c'b  -  cT)        '     '     ^^' 


_  {a"b  -  ab")  {a'd  -  ad')  -  (a'b  -  ah')  (a"d  -  ad") 
~  {c'a   -  ca')  {b"a  -ba")  —  {c"a-  a"c)    {b'a  -  ba' ) 


^8) 


Formulas  (6),  (7)  and  (8)  enable  us  to  solve  all  groups  of  simul- 
taneous equations  of  the  form  of  (1),  (2)  and  (3).     Thus, 

2x  +  3y  +  43  =  29, 
3a;  +  2y  +  5z  =  32, 

4a:  +  3y  +  22  =  25  :  ^^ 

Vy  comparison  with  (1),  (2)  and  (3), 

o    =  2,        5=3,        c    =  4,         d    =29, 
a'  =3,         b'  =  2,         c'  =5,         d'  =  32, 
a"  =  4,         b"  =  3,         c"  =  2,         d"  =  25  : 
by  substitution  in  (6),  (7)  and  (8), 

=  (  g  -  12)  (145  -  128)  -  (15  -    8)  (58  -  100) 
'      (10  -  12)  (     6  -  12  )  -  ;  4  -  16)  (15  -  8) 

_  —  102  +  2!>4  _  192  _ 
""         12  +  84    ~  "90"  ~    ' 


220 


APPENDIX. 


y  = 


;iO  -  12)  (58  -  100)  -  (4  -  10)  (145  -  128) 
(TO^  12)  ("6  -  12  )  -  (4  -    0)  (   15  -  8     ) 

84  +  204  _  288  _ 


12  +  84 


96 


z  = 


_  (12-    6)  (87  -  64)  -  (9  -    4)  (116 -50) 
(10  -  12)  (  6  -  12)  -  (4  -  IG)  (     4  -  9  ) 


138-330       192 


+  12-60 


48 


=  4. 


1.  Given 


EXAMPLES. 

'     X  -{■    y  +  z  =       90 

2a;  —  Sy  =z  —  20   f^ 

.  2a;  -  42  =  —  30  J 


to  find  X,  y  and  & 


By  comparison  with  (1),  (2)  and  (3), 

a  =  1,  h  ■=.  1,  c  =  1, 
a'  =  2,  6'  =  -  3,  c'  =  0, 
a"  =  2,         b"  =       0,         c"  =  -  4. 

bj  substitution, 


d  =  90 
(i''  =  -  20, 
d"  1=  -  30 ; 


4  X  20  -  3  X  -  330       910       „, 
-  =  — . — :— ^— =  35, 

V  = =  —  =  30, 


2  = 


-2 

X 

-  4  +  6  X 

3 

-2 

X 

-330  +  6 

X  20 

-2 

X 

4  +  6x3 

2 

X 

200  -  5  X 

210 

—  650 


-2X-2  +  6X-5        —  26 


-  =  25. 


2.  Given 


X  -\-    y  -\-    2=6" 
a;  +  2y  +  3z  =  14 
I  3x  —  y    +  42  =  13 


to  find  x,  y  and  z : 


ADDITIONAL   EXAMPLES.  221 

>  y  comparison, 

a  =  I,  b  =  1,  c  =  1,  d  =z  Q, 
a'  z=zl,  b'  =  2,  c'  =Z,  d'  =  14, 
a"  =  3,         b"  =  -\,         c"  =  4,         d"  =  13  : 

oy  substitution, 

_5x4-ll  _2xll-4_o  _4x-8+lx5_„ 

"^-2x5-1   -^'     ^-2^5^-^'      ^  -  ■~2^^Z.-43r -'^ 

ELIMINATION  BY  THE  METHOD  OF  ARBITRARY  MULTIPLIERS. 

3.  There  is  a  method  of  elimination  by  means  of  arbitrary  quan- 
tities that  will  often  be  found  useful,  particularly  in  the  higher 
investigations  of  applied  mathematics.  It  consists  in  multiplying 
both  members  of  one  of  the  given  equations  by  an  arbitrary  quan- 
tity, then  adding  the  resulting  equation  to  the  second  of  the  given 
equations,  member  to  member,  after  which  such  a  value  is  to  be 
assigned  to  the  arbitrary  quantity  as  will  reduce  the  co-efficient  of 
the  quantity  to  be  eliminated  to  0, 

To  illustrate,  let  us  take  the  two  simultaneous  equations, 

ax   -\-  by   —  c      ...      (1), 
a'x  -\-b'y  =  c'     '     '     .     (2) ; 

multiplying  both  members  of  (1)  by  n,  which  is  entirely  arbitrary, 

we  have 

nax  +  nby  =i  nc     -     •     •     (3) ; 

adding  (2)  and  (3),  member  to  member,  and  factoring, 

(«a  -}-«')*+  ("^  -\-  b')  y  =  nc  -\-  c'     .     •     •     (4). 
If  it  be  required  to  eliminate  y,  place 

nb  -\-  b'  =  0  ;  .  • .     n  =  —  — ; 


222  APPENDIX. 


<ubstituting  this  in  (4)  and  reducing,  we  find 

~  T  *"  +  ''       b'c-  be' 


^  =  —6— —  =  ^6^-^     •     •     •     (5). 


a' 


If  it  be  required  to  eliminate  x,  place 

na  -\-  a'  =  0: 

a 

substituting  in  (4)  and  reducing,  we  find 

a' 

a  ac   —  a  c  , 

y=  —^—-  =  ^^73^  •  •  •  («>• 

6  +  6' 

a 

These  values  of  x  and  y  correspond  to  those  already  deduced  by 
previous  methods. 

As  an  example,  let  it  be  required  to  find  the  values  of  x  and  v 
from  the  equations 

32;-    y  =  5       ...     (1), 
Ta;  +  3y  =  33     •     •     •     (2)  ; 
multiplying  both  members  of  (1)  by  n, 

3na:  —  ny  =  5ii     •     •     •      (3)  ; 
adding  (2)  and  (3),  m.eniber  to  member, 

(3n  +  7)  a;  +  (3  —  n)  y  =  5«  +  33     •     •     .     (4) : 
1st.  Assume         3  — 7i  =  0;         .•.     «  =  3; 
substituting  in  (4),  and  reducing, 

15  +  33       „ 
X  =  — -—  =  3 

9  +  7 

7 
2d.  Assume         3n  +  7=:0;         .*.     nt— -; 

o 


ADDITIONAL    EXAMPLES. 


223 


substituting  in  (4),  and  reducing, 

35 


+  33 


y- 


^-l 


1.  Given 


EXAMPLES. 


2         6^ 


>      to  find  X  and  y. 


Multiplying  both  members  of  the  first  equation  by  n  and  adding  to  the 
second,  member  by  member, 


making 


malting 


n  =  —  -         and  reducing, 
2 

y  =  i2; 

n  =  -         and  reducing, 
o 

ar  =  8. 


y-2        , 
X  —  ^-— —    =  5 


2    Given 


4y 


7 
a;-f  10 


=  3 


>■      to  find  z  and  y. 


Reducing  and  transposing, 

7«  -  y  =  33     •     •     .     (1), 
12y  —  a:  =  19     •     •     -     (2)  : 

multiplying  by  n,  and  adding  and  factoring, 


224 


APPKKDIX. 


(7n  -  1)  a;  -  (n  -  12)  2/  =  33n  -f  19 : 


making 
making 

3.  Given 


w  =  -,       we  find       y  =  2  ; 
n  =  12,         we  find         x  =  5. 


< 


r  9    6 

a;         7 


>-      to  find  a;  and  y. 


J 


Multiplying  by  n,  adding  and  factoring, 


(9n  +  14)  -  +  (6n  -  6)  -  =  36»  +  10 ; 
X  y 


making 
making 


«  = — ,     we  have     -  =  3 : 

9'  y 


y  = 


n  =  1,  we  have     -  =  2 :       .-.  x  =.-■ 

X  2 


MISCELLANEOUS  GROUPS  OF  SIMULTAISTEOUS  EQUATIONS  OF  THK 

FIRST  DEGRER 


1.  Given 


_1_        1   _2^ 

3a;       5y  ~  9   I 


L  5i  "^  3y  ~  4  j 


>■      to  find  X  and  y. 


Combining  and  elimii.ating  — , 


X 


whence, 


\25       9/2/      45       12' 


1              do  1 OQ 

_  —  •  •        11  —    I  ^ "» 

y-64'         •    -^-^35 


ADDITIONAL   EXAMPLES. 


226 


Comoining  and  eliminating  -> 


whence, 


V9       25/  X      27 


_1_ 
l2' 


1 

X 


65 
192' 


—  ■*  65 


2.  Given 


3  +  23,  =  5 


2x  —  1 


-y +1=0 


to  find  a?  and  y  ; 


Clearing  of  fractions  and  reducing, 

a;  +  6y  =     15     •     •     . 
2a;  —  5y  =  —  4     •     •     . 
oombining  and  eliminating  x, 

17y  =  S4;         .-.     y  =  2; 
substituting  in  (1),  x  =  3. 


(1), 
(2): 


3.  Given 


y  -2  ^ 


4y =  3 


to  find  X  and  y. 


Clearing  of  fractions  and  reducing, 

.      7x  —  y  =  33     •     .     . 
12y  —  a:  =  19     •     .     . 
combining  and  eliminating  x, 

83y  =  166;         .-.     y  =  2; 
substituting  in  (1),  x  =  5 


(1), 

(2); 


236 


APPENDIX. 


4.  Given 


■     5a:- 

4 

6 

20- 

2y 

+  2v  =  24 


+  5x  =  40| 


to  find  X  and  y. 


Clearing  of  fractions  and  reducing, 

5a:  +  12y  =  148, 
25a:  -    2y  .-  182 ; 
combining  and  eliminating  y, 

155a;  =1240;         .-.     a;  =  8  ; 
substituting,  we  find  y  =  9. 


5.  G 


iven 


2^3  6 


2'^3 


«+i 


a:       2 

2  +  3  =  '» 


to  find  a;,  y  and  s. 


Clearing  of  fractions  and  reducing, 

Sx  +  2y+z  =  72     •     •     • 
-  a:  -f-  3y  -f  2z  =  48     •     •     . 
3a:  -f  2*  =  60     .     .     . 

combining  (1)  and  (2),  eliminating  y, 

llx  —  2  =  120     •     .     . 
combining  (3)  and  (4),  eliminating  2, 

25a:  =  300  ;  .  • .     «  =  l2  ; 

by  substitution  in  (3),  z  =  \2, 

(1),  y  =  12. 


(1). 
(2), 
(3); 

(4); 


u 


ADDITIONAL    EXAMPLES. 


227 


0,  Given 


2  ^  3  ^  7 

X       y       z 

3  +  l+2  =  »' 


X       y       z 

6  +  l  +  6='» 


to  find  X,  y  am    >. 


Clearing  of  fractions, 

21a:  +  14y  -f    6z  =  924     •     • 
10a;  +    6y  +  152  =  930     •     • 
x+    ^y+      2  =  114     •     . 
combining  (1)  and  (3),  eliminating  2, 

15a;  +  2y  =  240     •     • 
combining  (2)  and  (3),  eliminating  z, 

5x  4-  24y  =  780     •     . 
combining  (4)  and  (5),  eliminating  a;, 

70y  =  2100  ;  .  • .     y  =  30  ; 

from  (5),  a:  =  12  ;     from  (3),     2  =  42. 


0), 

(2), 

(3); 

(4); 
(5); 


r  2a;  -  3y  +  2z  =  13     • 

■     •     (1)^ 

7.  Given     - 

2w  —    a;  =  15 

2y+    2=    7 

•     •     (2) 
.     (3) 

^  5y  +  3v  =  32 

.     .     (4)J 

Combining  (2)  and  (4),  eliminating  v, 

lOy  4-  3a;  =  19     •     . 

•    (5); 

combining  (5)  and  (1),  eliminating  x, 

29y  —  6z  =  —  1     .     . 

•    (6); 

(2)       t<   find  ar,  y,  g 


and  V 


=2518 


^:tj 


ATPENDIX. 


combining  (6)  and  (3),  eliminating  z, 

41y  =  41 ;         .  •.     y  =  1 : 
by  successive  substitutions,        z  =±  5,     a;  =  3,     i;  =  9. 


^23_J_4>k 
X       y~  12       z 


8.  Given 


3      5_  19      4 

a;       2  ~  24  "^  y 

V.  y        2         8       a:  J 


I"      to  find  X,  y  and  z. 


Transposing,  reducing,  &c., 


2l+3i-4i  =  l 

X  y  z       2^ 

3i-4l  +  5i  =  i| 

X  y  z       24 

X  y  z       24 


•  •  • 


(1), 
(2), 
(3): 


eombining  (1)  and  (3),  also  (2)  and  (3),  eliminating  -, 

_10l+43l  =  15    .    . 

X  y      24 


10 
24 


J.  «  y 

oombining  and  eliminating  -, 

y        24 

substituting  in  (5),       x  =  6;     whence,     2  =  8. 

10+   Qy—  4x 


(4), 

(5); 


1        2 

-  =  -,     or     y=12; 


9.  Given 


6a:—  9y+  3 

126+  8a:-17y 
I,  100-12"^+'7^  ~  13  J 


3 
35 


to  find  z  and  y. 


ADDITIONAL   EXAMPLES.  ^29 

Clearing  of  fractions  and  reducing,  .Aj 

-2x+      3y  =  -      1     .     .     .     (1), 
262a:  -  233y  =       931     ••     •     (2). 

Combining  and  eliminating  x, 

160y  =  800;         .-.     y  =  5; 
by  substitution  in  (1),  x  =  8. 

(°^  +  '''  =  '^) 

10.  Given  •<  a(a  +  a:)_       >•     to  find  x  and  y. 

iW+J)-    ) 

Clearing  of  fractions  and  reducing, 

ax  -\-  by  =  c^     •     •     •     (1), 

ax  —  by  =  V^  —  a?'      •     (2) ; 
*>y  addition, 

^2  _|_  J.2  _  fj2 

2ax  =  52  _j.  ^2  —  a^  .  ^ .  ^     -J.  _ _^ . 

by  subtraction, 

a2  +  c2  —  52 


2by  =  c^  +  a^ —b^;  .'.     y  = 


26 


MISCELLANEOUS  EXAMPLES  OF  EQUATIONS  OF  THE  FIRST,  SECOND 
AND  HIGHER  DEGREES,  CONTAINING  BUT  ONE  UNKNOWN 
QUANTITY. 

1.  Given  3ar2  —  4  =  28  +  3.2^     ^  fi^d  zx 

transposing  and  reducing, 

x^=\Q;        .-.     a;=±4 

o    Given         ?^^  -  ^-^  =  117  -   bx\     to  find  x; 

00 


230  APPKNDIX. 

Qearingof  fractions,  transposing  and  reducing, 

a;2  =  25  ;         .  • .     a;  =  ±  5. 

3.  Given  x^  -\-  ab  =  5x^,     to  find  x ; 

transposing  and  reducing, 

^'^  =  V '         •   •     ^  ~  "^  2  '^' 

An-  X    +1         X    -1  7 

4-^^^^"         IF^^Tx- x^  VTx  =  ^^^^iz'    •'^^^'- 

Clearing  of  fractions, 
a;4  4.  i4a;3  _  j^4a.2  _  io22a;  —  3577  -  a;*  +  14^3  +  24a;2  _  1022« 
-If.  3577  =  7a;3  -  343a; ; 

transposing  and  reducing, 

21x3  _  1701a;,     or    a;2  =  81  ;         . '.     a;  =  ±  9. 

5.  Given  ,J^+^J^T7l^     to  find  a:; 


multiplying  both  members  by   -y/a;  +  2, 

^'  V'a;  -2 

multiplying  both  members  by   •y/a;  —  2. 

a;  _  2  +  z  +2  =  4  -v/a;2  —  4,     or     a;  =  2  v/a;2  —  4 ; 

squaring  both  members, 

o       16       16      „  A    ^ 

,»  =  4x2  _  16,     or     a;2  =  —  =  —  X  3  ;         .  • .     x  z=  ± -y/Z, 

6.  Given  x  +  -/ox  +  10  =  8,  ,0  find  a; ; 


ADDITIONAL    EXAMPLES.  '^1 


Transposing  and  squaring  both  members, 

5x+  10  =  64- 16a;  +  x«; 
whence,  x^  —  21a;  =  —  54  ; 


21  /  441       21       15 

by  the  rule,         ^  =  '2"'^V^^'*"~4~~'2^'2"* 

.-.     a;  =  3,         a:  =  18. 

7.  Given  5  ^^  +  7  \f^  =  108,       to  find  x : 

make  y^  =  y ;     whence,     \f^^-y^\ 

substituting  and  reducing, 

„      7  108 


5 

> 

^ 

7 
10 

± 

'108 
5 

+ 

49 
100 

4^ 
10' 

whence,  y  =  —  —  ± 

27 
.  • .     y  =  4     and     y  =  —  —  ; 

from  which,     a;=±  v'?/^=rt8     and     a:=±  VP=  ±  y  l  —  yj 

8.  Given  Zx^  +  10a;  =  57,       to  find  x. 

By  division, 

x'-  -^  ~  X  ■=.  19  ;     whence, 
o 


5  /,^       25  5       14 

.•.     a;  =  3     and     x  =  —  6^« 

9.  Given  (a?  -  1)  (^  -  2)  =  1,     to  find  x. 

Performing  indicated  operations  and  reducing, 

a;2  --  3a;  =  —  1  ; 


232  APPENDIX. 

115 

10.  Given  «  ^^  ~  o  ^  =  o'     **^  ^"^  ^' 

<S  o  o 

DividUig  both  members  by  — ,  or  multiplying  by  2, 


11.  Given 


x^  —  -X  ^=  -\     whence, 
o  4 

1      /5    r  1    7 

«  =  -=b\/-  +  -  =  -±-: 
3      V4^9       3       6' 


2X-10      ar  +  3       ^      ,    .    , 

— =  2,     to  find  X. 

8  —  X        X  —  2         ' 


Clearing  of  fractions, 

2x2  _  14a;  +  20  -  (5x  -f  24  -  x')  =  20x  -  32  -  gx* ; 

reducmg,  x-^ —  x  = —\     whence, 


39         /     28       1521       39  ±  31 


100  10      ' 

.  • .     X  =  7,        «  =  p- 

0 

12.  Given — -  =  — -,     to  find  *. 

X  —  1       X  +  3      35 

Clearing  of  fractions, 

35  (x  +  3  —  X  +  1)  =  x2  _|.  2a;  —  3  ; 
reducing,  x^  +  2x  =  143  ;     whence, 

X  =  -  1  ±  -y/Iil  =  -  1  rb  12 ; 


ADDITIONAL   EXAMPLES.  335 

.-.     a;  =  11,         x=  -13. 

24 

13.  Given  x  -\ =  3r  —  4,     to  find  x.    . 

X  —  I 

Gearing  of  fract  ons, 

a;2  —  a;  +  24  =  3a;2  —  3a;  —  4a;  +  4 ; 
reducing,  a;^  —  3a;  =  10  ;     whence, 


'=W 


,^       9       3  ±7 
a;  =  5,         a;  =  —  2. 


ar  a;  +  1        13 

14.  Given  — -— r  H =  -r,     to  find  x. 

a;  +  1  a;  b 

Clearing  of  fractions, 

6x2  _|_  6x2  +  12a;  +  6  =  13x2  ^  13a. . 
reducing,  x^  +  x  =  6  ;     whence, 

5 


X 


-l-^f^,  =  -. 


2    : 

.  • .     X  =  2,         X  =  —  3. 

a;  —  4 
15.  Given  ~  =  a;  —  8,     to  find  x 

2  + yx 

Since  x  -  4  =  (y^— 2)   (y^  +  2), 

we  have,  by  performing  indicated  operations, 

yx"— 2  =  a;  —  8,     or     -y/x"=  x  —  6; 
squaring  both  menr  bers, 

X  =  x2  —  12x  +  36; 
or,  x2  —  1 3x  =  —  36  ;     whence, 

I 


234  APPENDIX. 

.  • .     X  —  0,         z  =■  A. 

16.  Given  17a;''!  +  19a;  -  1848  =  0,     to  find  z. 

,       19  1848 

Reducing,  z^  -\-  —  x  =  — -  ;     whence, 


19  _^      /1 848 

~  ~34~  V  "TT 


848        361        —  19  ±  355 


17     '    1156  34 

,•.     x  =  9\j,     and     a;  =  —  11. 


1  5 

17.  Given  -  a;2  -j-  -  a;  =  27,     to  find  x. 


Multiplying  both  members  by  3, 


a;^  +  —  a;  =  81  ;     whence, 


-  15  ±  39 


=  -T-A^  =  - 


.  • .     X  =1  6,     and     x  =  —  13  ^* 


18.  Given  a;  +  4  4-\/ ;  = -,     to  find  r. 

V  a;  —  4       X  —  4. 

Transposing  and  reducing, 

/^"4       28  -  a;2  /     ,    /      28  -  x^ 

\  / ;  = —  ;     or,     V  ^  +  4  =  — ,  : 

Va;  — 4        a;  — 4'         '      ^  -y/J^^ 

squaring  both  members  and  clearing  of  fractions, 

x2  —  16  =  784  —  56a:2  _|_  ^.i . 
reducing.  x*  -  57x2  =  _  800  ; 


ADDITION  Ax.   EXAMPLES.  235 

by  Rule,  Art.  124, 


/57  /      7Z      3249  /S7'      7 

hence,  .r  =  rt  5,     and     a;  =  ±  4-v/^- 


V 


2a;  +  9       4a:  —  3       „      3x  —  16      ,    .    . 
19.  Given  -^  +  ^_p3  =  3  +—3—,     to  find  a:. 

Clearing  of  fractions, 

16x2  ^  84a;  +  54  +  72a;  -  54  =  216a;  -h  162  +  12x2  _  55^;  _  48 ; 

5  114  , 

reducing,  x"^  —  -x  =  —;     whence, 

5  ±  43 


5^      /114   ,   25 


8       ' 
a;  =  6,     and     x  =  —  4 1- 


20.  Given         x^  +  x  +  2  ^x^  +  x  +  4  =  20,     to  find  x. 
Making  x^  +  x  =  y,         and  reducing, 


2  /y +  4  =  20  -  3/ ; 
squaring  both  members, 

4y  +  16  =  400  -  40y  +  y^ ; 
reducHig,  2/^  _  44y  =  -  384  ;       whence, 


y  =  22  ±  ^-384  -t-  4S4  =  22  ±  10 ; 
.-.     3/  =  32,     and     y  =  12  : 
Uking  the  first  value  . '  y  and  substituting  in  the  equation, 

x"^  +  X  =  y, 
x"^  +  X  =  32 ;     whence. 


236  ,?:  APPENDIX. 

taking  the  second  value  of  y, 

x"^  -\-  X  =  12]     whence, 


1  A„    ,    1       -  1  ±7 


21.  Giveh  \/x h  \/ 1 =  X,     to  find  z. 

transposing, 

squaring  both  members, 

X =  2;2  _  2  -^/^^^"^  +  1  _  2  . 

X  ^  X 

■whence,  by  reduction. 


a:^  —  a:  4-  1  =  2  \/x'^  —  a;. 
Placing  a;2  — a;^:?/     •     •     •     •     (I), 

y+l  =  2v^ 
squaring  both  members, 

y2  -(-  2y  +  1  =1  4y ;     whence, 

y2  -  22/  =  -  1  ;         .  • .     y=\±  ^-l  +  \,     or     y  =  1 ; 
substituting  in  (1),         x^  —  x  =  1  ; 

1  7^;        r       1  rb  -v/5~ 

22.  Given  a;^  —  63;  =  6ii  +  28,     to  find  a; :      ».' fi-;*  • 

transposing,  x"^  —  12a;  =  28  ;     whence. 


X  =  6  ±  V'28  +  St)  =  G=b8;         .'.     a;=14,     «=— 2. 


ADDITIONAL   EXAMPLES.  237 

23.  Given  a:*"  —  2x^'*  -f-  «»  —  6  =  0,     to  find  « : 

making,  x«  =  y  ;     whence, 

y4  _  2y3  4-  y  _  G  =  0  ; 

causing  the  second  term  to  disappear  (Arts.  263  and  313), 

3         91 
2"       10 

By  the  rule  for  solving  trinomial  equations  (Art.   124). 


'V4^16       16V       4  2^ 

.  • .     2  =  ±  ^  03,     and     2  =  ±  2  V""^ 


1    ,                                   IzhyOS           ,              l±y/-l 
bit,         y  =  -  +  z;         .'.     y= ^—,     and     y  = 1 


- —  "  /i  ±^13      "  /r±v^^ 

also,         x  =  Vy;         •••     *=\/ 2^ ^'     "^^V 2 

3;4   _L   2r3  -I-   8 

24.  Given  ,  T    '  -  \  ■  =  x^  +  x  +  8,     to  find  x, 

X^  -\-  X  —  o 

Clearing  of  fractions, 

T*  +  2x3  +  8  =  X*  +  2x3  +  3a.2  ^  2x  -  48 ; 

2  56 

reducing,  x"^  +  -  x  —  -—  ;     whence, 

o  o 


1  /56   ,    i        -1  db  13 

.  • .     X  =  4,     and     x  =  —  4^. 

2 
25,  Given  x^  —  1  =  2  +  -,     to  find  x. 

X 

Reducing,  x3  -  3x  -  2  =  0     •     •     •     (1) ; 


■2^3  APPENDLS. 

comparing  with  x^  -{-  px  -i-  q  =  0, 

p=-S,        q=-2; 
by  Cardan  s  formula, 

^  =  yr  +  \/l' = 2  : 

dividing  both  members  of  (1)  by  x  —  2, 

3;2  4-  2z  +  1  =  0 ;     whence, 

x=  -1, 

and  the  two  roots  are  each  equal  to  —  1  ;  hence,  the  three  roots 
are?   +2,   —  1,  and   —  ]. 

26.  Given  2x^  +  34  =  20x  +  2,     to  find  x. 
Transposing  and  reducing, 

a;2_  10x=  -  16;         .-.     x  =  S,     x  =  2. 

27.  Given  x^  =  56x  ^  -\-  x^,     to  find  x. 

multiplying  both  members  by  a:  ,  and  reducing, 

x^  -x^=bQ: 
cx)mparing  with  trinomial  equations  (Art.   124),  we  find 

3  J     1       2 

n  =  -,     and     -  ^  -  ; 
2  n      3' 

hence,  by  rule, 

2 

taking  the  upper  sign,         x  =  8^  =  4, 

•  '     lower     "  X  =  {-ly  =l/^. 

28.  Given         x^  —  \2xi -\- Ax  +  207  =  0,     to  find  x. 


ADUrnOKAL    EXAMPLES.  239; 

A  superior  limit  of  the  positive  roots  is  13  (Art.  279);   a  supe- 
rior limit  of  the  negative  roots  (numerically),  is  — 7  (Art.  281). 
By  the  method  of  Art    285,  rejecting  +  1  and   —  1,  which  are 
not  roots,  we  find  , 


DiTisors, 

23, 

27, 

3, 

-9, 

-1, 

0; 

3, 

69, 
73, 

-    3 

-69 
-65 

hence,  9  is 

a 

commensurable 

root. 

Dividing 

both  members 

by 

X  — 

9, 

we  find 

a;2 

-3x 

•         X  : 

-23  = 

^0 

;     or, 

a;2  -  3x  =  23  ; 

23  4- 

9 

_  3  ±  v'lOl 

29.  Given  x^  -f  Sx'^  —  6x  —  8  =  0,     to  find  x. 

This  is  solved  in  a  manner  similar  to  the  preceding. 

A  superior  limit  of  positive  roots  is  4,  and  of  negatiye  roots 
(numerically),  —  7. 

Divisors             4,            2,          1,  —  1,  _  2,  —  4, 

-2,-4,-8,  +8,  +4,  t-2, 

_S,     -10,  -14,  +2,  -2,  -4, 

_  2,     -    5,  -14,  -2,  +1,  +1, 

+  1,     -    2,  -11,  +1,  +4,  +4, 

-    1,  -  11,  -1,  -2,  -1, 

0,  -10,          0,  -  ],          0; 

hence,  x  =  2,         a?  —  —  1,         x  =  —  4, 

10 


240'  APPENDIX. 

30.  Given  x^  +  9.r  —  1430  =  0,     to  find  x. 

By  the  same  rule  as  before,  we  find  14  for  a  superior  limit  of  the 
real  positive  roots,  and  from  Art.  283  we  see  that  the  equation  has 
DO  real  negative  roots.     By  the  rule  (Art.  285),  we  have 

13,  11,  10,  5,  2,  1, 

-  110,     -130,     -143,     -286,     -715,     -1430, 

-  101,     — 121,     -  134,     -  277,     -  700,     -  1421, 

-  8,     -    11,  ...,  ...,       -353,     -1421, 

...,  0,  ...,  ...,  ...,      -  1420, 

hence,  11  is  the  only  commensurable  root. 
Dividing  both  members  by  a;  —  11,  we  have 

a;2+  lla;=  -  1.30: 


2        V  4  2 

28 

31.  Given  '\/x-\-  s^/x  +  l  = ,     to  find  x. 

V^  +  7 
Clearmg  of  fractions  and  transposing, 

Va:2  +  7a;  =  21  -x; 
squaring  both  members  and  reducing, 

49x  =  441  ;         .-.     a;  =  9. 

32.  Given  -^/a  +  a;  —  ^/a  —x  =  -y/^,     to  find  x. 
Squaring  both  members  and  reducing. 


2a  —  aa?  =  2 -/a*  —  a;* ; 
squaring  both  members, 

4a'  —  4r/2ar  +  a^x"^  =  4a2  —  4a:»  ; 


ADDITIONAL    EXAMPLES.  241 

reducing  and  dividing  both  members  by  a;, 
a:  (a2  +  4)  =  4a2 ;  .  • .     x  z= 


a2  +  4 

33.  Given         -^/Aa  +  x  =  2  ^b  +  x  —  ^/x,     to  find  r. 
Squaring  both  members  and  reducing, 

(a  —  b)  —  X  =  —  yjbx  +  a;2  ; 
squaring  both  members, 

{a  —  by  —  1{(i  —  b)x  -\-  a;2  =  ia;  -f-  a;2 ; 

hence,  (2a  —  6)  x  =  (a  —  6)2  ;         .  • .     x  =  ^"  ~    -^  . 

2a  —  b 

34,  Given        V''*^  +  x  +  -/a  +  a;  .—  2  -y/a:  —  2a,     to  find  * 
Squaring  both  members  and  reducing. 


2  ^\o?  -f  5ax  +  a;2  =  2z  -  13a ; 
squaring  both  members, 

16a2  +  20aa;  +  4a;2  =  4a;2  -  52aa;  +  169a2 ; 

reducing,  72oa;  =  153a2;         ,•.     x  = 

8 

1    -f-  2:3  1    —  a;3 

35.  Given  + =  a,     to  find  x. 

(1  +  ar)2        (1  —  xf 

Dividing  both  terms  of  the  first  fraction  by   (1  +  a-),  and  of  th« 
second  by  1  —  a;,  we  have 

1  —  a:  +  ar2       \  Ji^  x  ^  x^ 

1  +  «  \  —  x  ' 

dearing  of  fractions, 

1  -  2ar  +  2a;2  -  2;3  +  1  +  2x  +  2ar2  +  a:^  =  a  -  ax"^ ; 


reducing,         (4  -}-  a)  a;2  =  a  —  2  ;         .  • .     ar  =  ±  v  /- 

V  a 


-2 


343  APPENDIX. 


MISCELLANEOUS   EXAMPLES    OF   SIMULTANEOUS    EQUATIONS    OP 
THE   SECOND   AND   HIGHER   DEGREES. 

1     ...  ix"  +  y^  =  \^xy     ...      (1)  I     ,     -    ,  , 

J.  (jrivei)       <  V    to  find  x  and  y. 

U   +  y    r=  12        .     ,     .     (2)  > 
Make  x  =  v  +  w,     and     y  =  v  —  w. 

From  (2),  we  have     {v  -\-  w)  -\-  {v  —  lo)  —  \^ ;         .  • .     v  =  6. 
Frona  (1),  we  have,     (y  +  wY  +  (v  —  w)^  =  18(^2  —  w^) ; 
or,  reducing,  v'  +  Svii'^  =  9  (v'  —  e*'^) ; 

substituting  the  value  of  v,  vr  ^  have 

216  +  18w2  =  9  (36  -  w^),     or     27^2  =  108  ;         .  • .     w  =  ±  2  , 
hence, 
t=v,-|-tt'=:6±2  =  8     a^l     4;  y=:v  —  «'=  G=p 2  =4    and    8. 

ix^  +  y'-z53     .     .     .      (1)) 

2.  Given  •{  }•   to  find  x  and  y. 

(  -v  =  14     ...     (2)) 

Multiplying  both  menr>W.rs  of  (2)  by  2,  and  adding  and  subtracting, 
we  have 

a:2  +  2a;y       "2  =  81  ;         .  • .     x  +  y=  ±  9, 

x^  —  9rr,   r  y^  =  '^^  ;         • ' .     «  —  y  =  ±  5  ; 

nenc^,         x  r:.    >    H,     and       —  7      y  =  +  2,     and     —  7. 


^x-  +  7/*  =  82     .     .     .     (1) 
3.  Given  -j  }■   to  find  x  and  w. 

(r.    +y    =    4     .     .     .     {2y 


\ 


Raising  both  ro^nbers  of  (2)  to  the  4th  power,  adding  to  (1),  mem 
her  to  member,  and  dividing  by  2, 

X*  +  2a;3y  -|-  Sx^y^  +  2xv^  +  y*  =  169  ; 


ADDITIONAL   EXAMPLE8,  243 

extracting  the  square  root  of  both  members, 

x^  +  xy  +  y^=lS     .     .     .     (3); 

squaring  both  members  of  (1), 

x»  +  2a:y  +  y2=i6  .     .     .     (4); 

subtracting  (3)  from  (4),  member  from  member, 

xy  =  3,     or     3xy  =  9    .     .     (5) ; 
subtracting  from  (3),  member  from  member, 

x^  —  2xt/  +  t/2  =  4 ; 

whence,  x  —  y  =  ±2     .     .     .     (6). 

Combining  (2)  and  (6), 

X  r=S,     and     1 ;         y  =  1 ,     and     3. 

(5x   +3y    =r  19     .     .     (1)1  » 

4,  Given  <  >   to  find  x  and  v 

(7a:2 -2i/2  =r  10     .     .     (2)) 

From  (1),  we  find 

_  19  —  3y  2  _  3^1  -  114y  -\- 9y^ 

''-        5        '         •*•     *    -  25  ■ 

Substituting  in  (2),  and  reducing, 

798  2277 


y^ = 


13  13    ' 


399   ,       /-2277   .    159201       399  ±  360 


399   ,       /-2277 
whence,  y  =  _±^__  + 


109     ~         13 


hence,  y  =  — ,     and     y  =  3  ; 

1.    •      •                        40r>  ^ 

b_v  substitution,       x  z= — -,     and     a:  =  2. 


^44  ■■  ,.:•      APPENDIX. 

(x    +4y  =  14     .     .     (1)) 

5.  &i>*>.i  "i  >   tv  find  X  and  y. 

(y2  +  4a;  =  2y  +  11     (2)) 

From  (I),         V'  =  14  —  4y,     or     4a;  =  56  —  16y ; 
subtracting  ana  reducing, 

y2  _  I8y  =  -  45 ; 


whence,     y  =  U  ±  -y/—  45  +  81  =  9  ±  6  =  15      and     3 ; 

hence,  x  =  2      and     —  46. 

(  a:2  4.  4y2  ^  256  -  4ary     .     .      (1)) 
6.  Given        \  >■  to  find  a;  and  y. 

{  33/2 -x2  =  39     ...     (2)) 

Transposing  in  (1),  and  extracting  the  square  root  of  both  mernbera, 

a;-f2y=d=16;         .  •.     a;  =  ±  16  —  2y ; 
or,      •  «*  =  256  ^^  64y  +  4y2  ; 

substituting  in  (2),  and  reducing, 

y2  ::p  64y  =  —  295  ; 


whence,      y  =r  ±  32  =b  -/  -  -^5  -|-  1024  =  ±  32  dr  27 ; 

.  • .     1/  =  ±    59     and     ±  5  ; 

substituting,  z  =  ±  102     and     ±  6. 

ix^-    y^  =  24     .     .     (1)) 

7.  Given  \  >  to  find  x  and  y. 

(a;2  +  a:y   =84     ..     (2)) 

Subtracting  (1)  from  (2).  member  from  member, 
y2  4-  ary  =  60     .     .     .     (3)  ; 
adding  (3)  to  (2),  member  to  member, 

a:*  -♦-  2a;y  +  y2  _  144 .  , .,     x  +  y  =  ±  \2     .     .     .     (4). 


ADDITIONAL    EXAMPLES. 

Dividing  (1)  by  (4),  member  by  member, 

x  —  y=±2; 
hence,  x  =  d=  7,         y  =  ±  5. 

-v=    4     .     .     .     (1) 
xy  =  45     .     .     .     (2) 

From  (1),  a;  =  y  4-  4,     which,  in  (2),  gives 

y»  +  4y  =  45  ; 


245 


8.  Given 


ix-y  = 
{       xy  = 


to  find  r 


.-.     y  =  -2±  -/45  +  4  =  -  2  ±  7 ; 
.  • .     y  =  +  5,     and     —  9  ; 
and  by  substitution,        x  =  -^  9,     and     —  4. 

ixy-\-xy^  =  l2     ...     (1) 
9.  Given  < 

(    X  +  xy3  =18     .     .     .     (2) 

Dividing  (2)  by  (1),  member  by  member,  and  reducing, 

1  +y^  _  3  l-y  +  y^  _  3 

y{l+y)      2'    '''*  y  ~  2' 


to  find  X  and  y. 


y''  -  2  y  =  - 1 ; 


by  reduction, 

I-  5  ^      /      ,    ,   25       5       3 

whence,  y  =  _  ±  ^  _  1  +  _  =  _  ±  _. 


• .     y  =  2,     and     y  =  - ; 


by  substitution, 


10.  Given 


X  =  2,     and    x  =.  16. 
.     (1) 


^1        1 

a;        y 

I  a;^      y^ 


(2) 


to  find  z  and  y. 


246  APPENDIX. 

From  (1),  we  find, 

1  1  1  2       2a   ^  1 

-  =  a ,     or     —  =  a2 b  -^i 

y  X  y^  X       x* 

substituting  in  (2),  and  reducing, 

'\\       b  -a^ 


e;-"©= 


2       ' 


1        a  lb  —  a^       a^       a  ±  J\ 

z      2      V      2       ^4  \ 


2b  —  a^ 

2  ' 


by  substitution,  -   =: — • 

2  ^  9 

hence,  x  ■=. ,     and     y  = 


a  ±  y/2b  -  a?-  a  =f  ■v/26  —  a^ 

rz^      4^^85  ) 

11.  Given  ly'^y        9  ^  ^  I    to  find  ar  and  y. 

(    a;  -  y  =  2         •     •      (2)  j 

Clearing  (1)  of  fractions, 

9x2  _|_  36a.y  _  85y2     .     .     .     (3). 

F^om  (2),         ar=:2  +  2/;         .-.     a;2  =  44-4y+y'; 

substituting  in  (3),  and  reducing, 

2_27      ^^ 
^        10  ^       10' 


27  / 

whence,  y  =  —  ±  y/ 

.  • .    y  =  3     and    y  =  — 

by  substitution,  a;  =  5     and     a:  = 


27   .       /_9_      729  _  27  ±  33  ^ 
To  "*"  400  ~       20     ■  * 

3^ 

10 

17 
10* 


ADDITIONAL    KXAMPLK8.  247 

X^         y^         X        y  _  27 

12.  Given      X^^  ~i^      y       x~  ~i  '  '  ^'  \  to  find  x  and  y. 


x-y  =  ^ (2) 


^  ,  y 


Make         -■\-~—z    (3) ;     whence,  by  squaring,  &c., 
y       X 

^  +  ^  =  ^'-2; 

yi  X^ 

hence,  from  (1),  by  substitution  and  reduction. 


z^  -k-  z  —  -r\ 


35 
4 


_       i  /35       1  _  -  1  d:6 


^         ^  "^ 

or,  2  =  2     and     -  - ; 

substituting  the  positive  value  of  z  in  (3),  and  clearing  of  fractions, 

5 

a;2  +  y2  _  _  a-y     .     ,     .     (4)^ 

i>om  (2),         a;  =  y  +  2  ;         .  • .     a;2  ^  y2  +  4y  +  4, 
and  a;y  =  y2  _j_  2y  ; 

substituting  in  (4)  and  reducing. 


2/24-2^  =  8;         .-.     y=  —  1±  v^8  +  1  =  —  1  dt  3  ; 
hence,  ?/  =  2     and     y  =  — -  4  ; 

by  substitution,  a;  =  4     and     y  =  —  2. 

{a;2  +  y^  +  s2  =  84     •     •     •     (1)  >| 
X   -{-  y    +  z   =  14     •     .     .      (2)    I     to  find  ar,  y, 
:r2  =  y2  •     •     •      (3)  J  ^"<^  ^• 

Substituting  in  (1)  and  (2)  the  value  of  y  from  (3),  and  reducing 
r2  4-  2ar2  +  22  =  84  +  rz     or     x  -\-  z  =  -^84  +  xz     •     •     (4). 

From  (2)  and  (3),  x  +  z  =  U  —  ^xz     -     •     (5). 


248  APPENDIX. 

Equating  the  second  members, 

14  -  y^  —  ■/84  +  xz  ; 
squaring  both  membeis, 

19G  -  28  -y/rz  +  xz  r=  84  +  arz ; 
•oducing,  y'^  i- 4,     or     xz  =  IQ     •     •     •     (0) ; 

h^nce,  from  (5),  x  +  z  =  10     •     .     •     (7); 

•obstituting  in  (7)  for  z  its  value  — ,  and  reducing, 

x^  —  lOx  =  —  16 ; 


«^nulCx^  a;  =  5±  -\/—  lt5  +  25  =  5±3; 

or.  X  =  8,         X  =  2: 

by  substitution,     z  =  2,     z  =  8,     and     y  =  ±  4. 


14.  (jriven     ■{  „         „  .       ?■   tt>  find  x  and  y. 

(  a:2  +  y2  ^  41  •  .  (2)  >  ^ 

From  (1),  by  transposition, 

^x  +  y  =  l2-{x  +  y); 
squaring  both  members, 

a:  +  y  =  144  -  24  (x  +  y)  +  (a:  +  vY ; 
reducing,  {x  +  y^  —  25  (a;  +  y)  =  —  144 ; 


25  /      , ,,    ,    025        ,  25  ±  7 

.-.     ^-i-y  =  -t-_±Y/-  144  +  — =  +— g— ; 

whence,  a;  -|-  y  =.-  16,     or     z  +  y  =  9. 

The  first  value  does  not  satisfy   (1),  unless  the  radical   have   the 
negative   sign  ;    adopting,  therefore,  the   second   value,  from   which 

X  =  9  —  y,     or     a;^  =  81  —  18y  +  y^, 


ADDITIONAL    EXAMPLES.  349 


which  in  (2)  gives,  after  reduction, 

y^  —  9j/  =  —  20  ;     whence, 


y 


=  -dz^^-20  +  —  =  —^^,        .-.     y  =  5    and    y  =  4; 


by  substitution,  x  =  4     and     x  =  5. 

lx^-y^=m     '     •     •     (1)) 
15.  Given       \  }■    to  find  x  and  y. 

(x   -y   =      3     •     .     .     (2)) 

Cubing  both  members  of  (2),  subtracting  from  (1),  member  from 
member,  and  dividing  both  members  by  3,  we  have 

xhj  —  xy^  =  30,     or     (r  —  7/)  xy  =  aO     •     •     •     (3)  ; 

dividing  (3)  by  (2),  member  by  member, 

a;y  =  10;         .-.     y  =  —', 
substituting  in  (2),  and  reducing, 

x^  —  3x  -10; 


3  /  9       3       7 

whence,  x  = -±^^ \0  + -  = -± -; 

•     •       X  ^  D,  X  n^    —  ^  1 

by  substitution,  y  =  2,     and     y  =  —  5. 


0^2 -f  ryxf—    208     .     •     (1)  ) 
16.  Given     \  V  to  find  x  and  y 

y2  4-'y-y^=1053      •      •      (2)) 
These  equations  may  be  written, 

642  4     1         2 

z^-fa;^    =    208,     or     o^{^^if)^    208     ...     (3), 

/  -f-  y^}  =  1053,  y^(y^  |-  x^)  =  1053     •     .     •     (4), 


250  APPJiNDIX. 

Dividing  (3)  by  (4),  member  by  member, 

x^       208        16 


I       1053       81 
extracting  the  4th  root  of  both  members, 


(5); 


y 

substituting  in  (3), 


\~  '6'  •    •      ^  -3      ' 


3/^  +  —  2/^  =  208  ;     whence, 


729  ^     '    81 

208 

—  3/2^208,     or     2/2  z=  729;         .-.     y  =  ±  27, 

and  by  substitution,  ar  =  ±    8. 

MISCELLANEOUS   PROBLEMS. 

1.  A  courier  starts  from  a  place  and  travels  at  the  rate  of  4  miles 
per  hour ;  a  second  courier  starts  after  him,  an  hour  and  a  half 
later,  and  travels  at  the  rate  of  5  miles  per  hour  :  in  how  long  a 
time  will  the  second  overtake  the  first,  and  how  far  will  he  travel  ? 

Let       X     denote  the  number  of  hours  travelled  by  2d  courier  : 
then  will  x  -\-  \^     " 
bx  « 

and        A{x  +  \\)    " 

From  the  conditions  of  the  problem, 

S-r  =r  4  (x  +  li)  ;         .  • .     x  =  Q>     and     5x  =  30. 

2.  A  pel  on  buys  4  houses  for  $8000;  for  the  second  he  gave 
half  as  much  again  as  for  the  first ;  for  the  third,  half  as  much  again 
as  for  the  second  ;  and  for  the  fourth,  as  much  as  for  the  first  and 
third  together  :  what  does  he  give  for  each  \ 


u 

(( 

1st 

(( 

"  miles 

« 

2d 

u 

(1      (( 

(( 

1st 

.( 

ADDITIONAL    EXAMPLES.  251 

Let     X  denote  the  amount  paid  for  1st  house  ;  then  will 

X  4-  -  "       "         "  "      •'    2d       " 

2 

2x  +  ~         "       "         "  "      "    3d       ** 

4 

3ar  4-  -  "       "         "  "      "    4th      " 

4 

From  the  conditions  of  the  problem, 

8a:  =  8000;  .'.      a:  =  1000 

a:  +  ^  =  1500 

2ar  +  ?  "  2250 

3x  +  ?  =  3250 
4 

3.  A  and  B  engaged  in  play  :  after  A  had  lost  $20,  he  had  one 
third  as  much  as  B  ;  but  continuing  to  play,  he  won  back  his  $20, 
to^^ether  with  $50  more,  and  he  then  found  that  he  had  half  as  much 
again  as  B  :  with  what  sums  did  they  begin  ? 

Let  X  and  y  denote  the  sums  with  which  A  and  B  began. 

Then  from  the  conditions, 

y-f  20 

a;  +  50=  (y  -50)  X  H; 
i^hence,         3a;  -    60  =    y  +    20 ;         .-.     a;  =    70 
2z  +  100  =  3y  -  150  ;  y  =  130. 

4.  A  can  do  a  piece  of  wort    in  10  days,  which  A  and  B   ogethot 
can  do  in  7  days :  in  how  many  days  can  B  do  it  alone  ? 
Let  X  denote  the  number  of  days. 


252  APPENDIX. 

Since  A  and  B  togethe    can  do  i  of  the  v>  i>rk  in  1  day,  A  can  do 
jJj  of  it,  and  B,  -  of  it  in  1  day  ;  hence,  from  the  relations  existing, 


X 

J_       1_1  ,       1_^ 


7a  "^  r  ~  o^  >  •  '  •       ~  —  r^Ti'       ^^       ^  —  ^^5- 


5.  A  person  has  $650  ir.vested  in  two  parts  :  the  first  part  draws 
interest  at  3  per  cent,  and  the  second  at  3^  per  cent,  and  his  total 
income  is  $20  per  annum  :  how  much  has  he  invested  at  each  rate  ? 

Let  X  denote   the  numler  of  dollars  at  3  per  cent:    then  will 


650  —  X  denote  the  number  at  3^. 


From  the  conditions. 


3x        650  —  X 

H ^;^ —  X  3;  =  20 


100     100      2     5 
whence,        Sx  +  2275  —  S^x  =  2000 ; 

.  •.  ^  =  275,  or  a:  =  550  :    .  • .  650  -  a;  =  100. 
2  ' 

6.  A  boatman  rows  with  the  tide,  in  the  channel,  18  miles  in  1-^ 
hours ;  he  rows  near  the  shore  against  the  tide,  which  is  then  only 
three-fifths  as  strong  as  in  the  channel,  18  miles  in  2i  hours  :  what 
is  the  velocity  of  the  tide  per  hour  in  the  channel  1 

Let     X     denote  the  velocity  of  the  tide  in  the  channel : 

Sx 
then,     —  "      "  "         "  "         near  shore ; 

5 

and       118 —  J  -r- 1^  will  denote  the  rate  of  rowing,  neglecting  tide 

also,     (l8  +  ^)^2J-      "         "  «  «        « 

hence,  (l8  -  _)  x  -  ^-    ^S  + —)  x  ^; 

Sx 
or,  i2-ar  =  8  +  — ,         ..     x  =  2^. 


ADDITIONAL    EXAMPLES.  253 

7,  A  garrison  had  provisions  for  30  months,  but  at  the  end  of  4 
months  tne  number  of  troops  was  doubled,  and  3  months  afterwards 
it  was  reinforced  by  400  troops  more,  and  the  provisions  were  ex- 
hausted in  15  months  :  how  many  troops  were  there  in  the  garrison 
at  first  1 

Let  X  denote  the  number  cf  men  at  first ;  then  will  30a;  denote 
the  number  of  months  that  one  person  could  subsist  on  the  provi- 
sions, or  the  number  of  month  y  rations  in  the  garrison. 

4x     denotes  the  number  of  monthly  rations  used  in  4  months, 
6x         "  "  "  "  "    the  next  3 

(2.r+ 400)8  "  "  "  "  "         "         8       " 

hence,       2Gx  +  3200  =  30j;,     or     4x  —  3200  ,       .'.  x  —  800. 

8.  What  is  the  number  whose  square  exceeds  the  number  itself 
by  6? 

Let  X  denote  the  number. 

From  the  conditions, 

1  ±5 


x^  —  x  =  Q;         .'.     a;  =  i±-v/6-|-i  = 
.  • .     a;  =  3     and     —  2. 

9.  Find  two  numbers  such  that  their  sum  shall  be  15,  and  the  sum 
of  their  squares  117. 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

X   -\-y    =    15     .     .     .     (1), 

x^  +  y^  =  m     .     .     .     (2). 

From  (1)         a:  =  15  —  y,     or     a-2  -  225  —  30y  +  y^  ; 


2o4  APPENDIX. 

subsfituting  in  (2)  and  reducing, 

y^  —  \hy  =  —  54. 


'•'>  /      7.    ,    225       15^3  ^         n 

J/  =  -2^y  -  54  +  —  =  —  ±  -,     or     y  =:9     and     x  ~  6, 

or     X  =  9     and     y  =  6, 

10.  A  cask  whose  contents  is  20  gallons,  is  filled  with  brandy  ;  a 
certain  quantity  is  drawn  off  into  another  cask  of  the  same  size, 
after  which  the  latter  is  filled  with  water :  the  first  cask  is  then  filled 
with  this  mixture ;  it  then  appears  that  if  6|  gallons  of  this  mixture 
be  drawn  from  the  first  into  the  second  cask,  there  will  be  equal 
quantities  of  brandy  in  each.  How  much  brandy  was  first  drawn 
off] 

Let  X  denote  the  number  of  gallons  first  drawn  off.  Then  will 
20  —  X  denote  the  quantity  remaining  as  well  as  the  quantity  of 

X 

water  added  to   the  second  cask  ;  — -  will   denote    the  quantity   of 

lirandy  in  each  gallon  of  the  mixture,  and 

X  x^ 

"^20'     "^     20 

will  denote  the  quantity  of  brandy  returned  to  the  first  cask,  which 
will,  therefore,  contain 

gallons  of  brandy.     Each  gallon  of  this  new  mixture  will  contain 
5>g  of  the  brandy  in  the  cask,  or 

400  —  20a:  +  x"^ 
400  ' 

nence.  6|  gallons  will  contain 

400  —  20j;  +  «» 
60 


ADDITIONAL    EXAMPLES.  255 

gallons;  and  after  this  is  drawn  oft',  10  gallons  must  remain  }  henoe, 
400  -  20a;  +  x^       400  —  20a;  -f-  x'^ 


20  60 

whence,  800  -  40a;  +  2a;2  =  600, 

or,  a;2  -  20x  =  100 ; 


=  10; 


.'.     a;=  10  ±  ^-  100+  100,     or     a;  =  10. 

11.  What  number  added  to  its  square  will  produce  42  ^ 
Let  X  denote  the  number. 

From  the  conditions  of  the  problem, 

a;2  +  a;  =  42  ; 

-  1  ±  13 


•.     a;  =  -  I  +  -v/42-t-i  = ^ 5     •  ••  x  =  6  and  ;r-  -7. 

12.  The  difference  of  two  numbers  is  9,  and  their  sum  multiplied 
by  the  greater  gives  266  :  what  are  the  two  numbers  ? 
Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

x-y  =  9     .     .     .     (1), 

x(x  +  y)  =  2GG     .     (2). 

From  (1),  y  =  X  —  '9;         substituting  in  (2), 

X  (2x  -  9)  =  266,     or     x^-^x=  133  ; 


9        I  sT" 

whence,  z=-±v/l33H = 

4      V  16 


81       9  ±47 
16 


.'.     X  =  14,         a;  =  —  9^; 
whence,  y  =  5,         y  =  _  18^. 

13.   A  person   travelled   105  miles:   if  he  had   travelled  2  milen 


11 


25<] 


APPENDIX. 


j)er  hour  slower,  he  would  have  been  6  hours  longer  in  completing 
the  journey  :  how  many  miles  did  he  travel  per  hour  ? 

Let  X  denote  the  number  of  miles  travelled  per  hour.     Then  will 

105 

denote  the  number  of  hours. 

X 

From  the  conditions, 
105         105 


a:  -2 
reducing, 


+  6,     or     105x  =  105x  -  210  +  Cz^  _  12a: ; 


a:2  _  2x  =  35  ; 
.-.     X  —\  ±i  -y/SS  +  1  =:  1  i  6  ;         .-.     a;  =  7. 

14.  The  continued  product  of  four  consecutive  numbers  is  3024' 
what  are  the  numbers  ? 

Let  X  denote  the  least  number. 
iVom  the  conditions  of  the  problein. 

x{x  +1)   (a;  +  2)  (a:  +  3)  =  3024, 
or  a:*  +  ^x"^  +  1  la:2  +  6a;  —  3024  —  0. 

A  superior  limit  of  the  real  positive  roots  is  9  (Art.  279).  Ne- 
glecting the  divisor  1,  and  all  negative  divisors,  we  may  proceed  by 
the  rule  (Art.  285),  as  follows  : 

9,  8,  7,  6,  4, 

-336,     -378,     -432,     —504,     -756, 
-330.     -372,     -426,     —498,      -750, 

-  83, 

-  72, 

-  12, 

-  6, 

-  1, 

-  0, 

Hence,  6  is  the  required  value  of  x,  and  the  numbers  are  6,  7,  8 
and  9. 


3, 

2, 

-  1008, 

-1512, 

-  1002, 

-  1506, 

-    334, 

-    753. 

-    323, 

-    742, 

•  • 
> 

-    371, 

"> 

-    365, 

••> 

•% 

ADDITIONAL    EXAMPLES.  251 

15.  Two  couriers  start  at  the  same  instant  for  a  point  39  miles 
distant ;  the  second  travels  a  quarter  of  a  mile  per  hour  faster  than 
the  first,  and  reaches  the  point  one  hour  ahead  of  him  :  at  wh.U 
rates  do  they  travel  1 

Let  X  denote  the  number  of  miles   per  hour  of  first   ooiirier. 

39 
Then  will   —  denote  the  number  of  hours  he  travels, 

X 

From  the  conditions, 

39       ,  39  „  39  1 

1  =  -— -r,     or     S9x  +  —  -x^--x  =  S9x: 

X  a;  +  1  4  4 

A     •  2.1  39 

reducme,  x^  A —  x  =  ■ — • ; 


8      V  ' 


-    ,       /39  ^    1         -  1  ±  25 


16.  The  fore-wheels  of  a  wagon  are  5^  feet,  and  the  hind-wb'iels 
1^  feet  in  circumference  ;  after  a  certain  journey,  it  is  found  thai  the 
fore-wheels  have  made  2000  revolutions  more  than  the  hind-whoels  • 
how  far  did  the  wagon  travel  ? 

Let  X  denote  the  number  of  feet. 

From  the  conditions  of  the  problem, 

•f         X  « ^  ^ 

^--=:-2000; 

multiplying  both  members  by      ^     , 

_  2394000  _  598500 

57  ar  -  42  a:  =  598500, 

15  x  =  598500, 

x=    39900. 


258  APPKNDIX. 

17.  A  wine  merchant  has  2  kinds  of  wine  ;  the  one  costs  9  shil- 
lings per  gallon,  and  the  other  5.  He  wishes  to  mix  them  together 
in  such  quantities  that  he  may  have  50  gallons  of  the  mixture,  and 
so  that  each  gallon  of  the  mixture  shall  cost  8  shillings. 

Let  X  and  y  denote  the  number  of  gallons  of  each,  respectively. 

From  the  conditions, 

j:  +    y  -  50      .     .     .     .      (1), 
0:r  +  52/  =  8(:r  +  y)     .      .      (2); 
Mubstituting  f^jr  x  -{-  y  its  value  in  (2), 

*dx  +  5i/  nz  400    .     .     .     .      (3)  ; 
combining  (1)  and  (3), 

4y  =  50;         .-.     y  =  12^,     and     x  =  Z1\. 

18.  A  owes  $1200  and  B,  $2500,  but  neither  has  enough  to  pay 
his  debts.  Says  A  to  B,  "Lend  nie  the  eighth  part  of  your  fortune, 
and  I  can  pay  my  debts."  Says  B  to  A,  •'  Lend  me  the  ninth  part 
.)f  your  fortune,  and  I  can  pay  mine :"  what  fortune  had  each  ? 

Let  x  and  y  denote  the  number  of  dollars  in  the  fortunes  of  A 
and  B. 

From  the  conditions  of  the  problem, 

a;  -I-  ^  =  1200,         or         8j;  +    y  =    9600, 

8 

/      2/  +  ^  =  2500,         or  a;  -f-  9y  =  22500 ; 

combining  and  eliminating  ar, 

71y  =  170400  ;         .  • .     y  =  2400,         x  =  900. 

19.  A  person  has  two  kinds  of  goods,  8  pounds  of  the  first,  and 
\i  of  the  second,  cost  together  |>18,46;  20  pounds  of  the  first,  and 
16  of  the  second,  cost  together  $36,40:  how  nmch  does  each  cost 
per  pound  1 


AODITIONAL    EXAMPLES.  259 

Let  X  and  y  denote  the  cost  of  a  pound  of  each  in  cents. 
From  the  conditions  of  the  problem, 
8a:  +    %=  1846, 
20x  +  16y  ==  3640  ; 
combining  and  eliminating  x, 

13y  =  1950:         .-.     y  —  150,     and     x  =  62. 

20.  What  fraction  is  that  to  the  numerator  of  which  if  1  be 
added  the  result  will  be  1  but  if  1  be  added  to  the  denominator  tha 
result  will  be  1-'^ 

Let  X  denote  the  numerator,  and  y  the  denominator. 

From  the  conditions  of  the  problem, 

*  "^  ^        ^  or         3a;  +  3  =  y, 

or  4a:  =  1  +  ?/ ; 

hence,  by  comliination,     a:  =  4     and     y  =  15.     Aas.  ^. 

21.  A  shepherd  was  plundered  by  ihree  parties  of  soldiers.  The 
Hrst  party  took  \  of  his  flock  and  i  of  a  sheep  ;  the  second  took  \ 
of  what  remained  and  ^  of  a  sheep ;  the  third  took  ^  of  what  then 
remained  and  ^  of  a  sheep,  which  left  him  but  25  sheep :  how  many 
had  he  at  first  % 

Let  X  denote  the  number  of  sheep.  Then,  after  being  plundered 
by  the  1st  party,  he  would  have 


y 

3 

X 

1 

1 

+  y 

4 

^-(4  +  4)==— 4—  '^^^P5 


after  being  plundered  by  the  2d  party,  he  would  have 
3a:  —  1        /3x  —  1    .    1\       x  —  \ 


1        /3x  -  1        1\ 
I     12       "^  3/ 


260  APPENDIX. 

after  being  plundered  by  the  3d  party,  he  would  have 
X  —  1        /x  —  1        1\       X— 3 


/x-l        1\ 


2 

from  the  conditions  of  the  problem, 

^ 3 

— ^—  =  25,     or     X  -  3  =  100  ;         .  •.     x  =  103. 

22.  What  two  numbers  are  those  whose  product  is  63,  and  the 
square  of  whose  sum  is  equal  to  64  times  the  square  of  their  dif 
ference  ? 

Let  X  and  y  denote  the  two  numbers. 
From  the  conditions  of  the  problem, 

xy  =  63 (1), 

{x-\-yy  =  64{x-yY     .      .     (2); 
extracting  the  square  root  of  both  members  of  (2), 

x  +  y  =  8{x  -y),     or     7x  =  9y;  .-.      a;  =  f  y  ; 

substituting  in  (1),  l^y^  =  6S; 

.  • .     y2  _  49     and     y  =  7,     also    x  z=  9. 

23.  The  sum  of  two  numbers  multiplied  by  the  greater  gives 
209  ;  their  sum  multiplied  by  their  difference  gives  57  :  what  are 
the  two  numbers  ? 

Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

(x -{- y)  X  =  209,     or     x^-{xy=209  .     (1), 

(x  +  y){x-y)==    57,     or     x^  -  y^  =    57     .     .     (2); 
subtracting  (2)  from  (I),  member  from  member, 

xy  +  y^=  152     .     .     (3) ; 
idding  (3)  and  (1),  member  to  member, 

a:2  +  2a;y  4- 2/2  =  361 ;         .-.     x -\- y  =  ]9  ; 


ADDITIONAL    EXAMPLES.  261 

209 

hence,  from  (1),  x  =  -— -  =  11 ;     also,     y  =  8. 

24.  Three  numbers  are  in  arithmetical  progression  ;  their  sum  is 
I ;"),  and  the  sum  of  their  cubes  is  495  :  what  are  the  numbers  ? 
Let  a    i/  and  z  denote  the  numbers, 

From  the  conditions  of  the  problem, 

y    —  X   =  z    —      y     .     .     (n 

X   +y    ^z   =     15     .     .     (2) 
a:3  +  y3  4.  ^3  _  495     _     _     ^3)  . 

trom  (1),         2y  =  z  -f-  ar,       which  in  (2),  gives       y  =i  5; 

substituting  in  (2)  and  (3), 

z   -\-   x=    10     .     .     (4) 
^3  +  a;3  =  370     .     .     (5)  ; 

dividing  (5)  by  (4),  member  by  member, 

z^  -zx  +  x^  z=S7     .     .     (6)  ; 

squaring  both  members  of  (4), 

z^  +2zx  +  x^  =  100     .     .     (7) ; 

combining  (6)  and  (7), 

'6zx  =  63,     or     zx  =  21 :         .  • .      2  =  — : 

X 

substituting  in  (4), 

21 
x-\ =  10,       or       a^  —  lOar  =  —  21  - 

X 


..     a; •=  5  ±  -v/—  21  +  25  =  5  ±  2 ;         hence,     «  =  7,     or     3. 

25.  Divide  the  number  16  into  two  parts  such  that  25  times  the 
square  of  the  first  shall  be  equal  to  9  times  the  square  of  Lhf 
seiJond. 


262  APPENDIX. 

Let  X  denote  one  part ;  then  will    \Q  —  x    denote  the  other. 
From  the  conditions, 

25x2  ^  9  (256  -  32a;  +  x"^)  =  2304  -  288a;  -|-  9x2 . 

reducing,  x"^  +  18a;  =  144  ; 


.  • .     a;  =  —  y  ±  '/144  +  81  =  -  9  ±  15,     or     x  =  6, 

since  the  negative  value  does  not  satisfy  the  problem  understood  in 
the  numerical  sense. 

26.  There  are  two  numbers  such  that  the  greater  multiplied  by 
the  square  root  of  the  less  is  18,  and  the  less  multiplied  by  the 
square  root  of  the  greater  is  12  :   what  are  the  numbers  ? 

Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

yV^=.18     .     .      (1) 
a;V7-12     .     .      (2); 
multiplying  (1)  by  (2),  member  by  member. 


■y/{xyf  =  216  ;  .  • .     xy  =i  m     .     .      (3)  ; 

adding   (1)  and    (2),  member  to  member, 

('v/^+  VF)  VV-SO     .     .     (4), 
or  -^/x  +  -y/v  =  5  ; 

squaring  both  members, 

a;  +  y  +  2v^'=25     .     .     (5), 
or  a;  +  y  =  13  .     (6); 

combining  (3)   and   (6), 

x  =  9.         V  —  4. 


ADDITIONAL    EXAMPLES.  263 

27.  What  two  numbers  are  those  the  square  of  the  greater  of 
which  being  multiplied  by  the  lesser  gives  147,  and  the  square 
of  the  lesser  being  multiplied  by  the  greater  gives  63  1 

Let  X  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

a;2y  =  147     .     .      (1) 
xy^  =    63     .     .      (2)  ; 
multiplying  (1)   and  (2),  member  by  member, 

xY  =  9261     .     .     (3), 
or  ^7  =       21      .     .     (4)  ; 

dividhig  (2)  by  (4),  member  by  member, 

y  =  S  ;         in  like  manner,         x  —  7. 

This  method  of  solution  might  be  applied  to  the  equations  of  the 
preceding  example. 

28.  There  are  two  numbers  whose  difTerence  is  2,  and  the  product 
of  their  cubes  is  42875  :  what  are  the  numbers  1 

Let  X  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

x-y  =  2     .     .     (1) 
xY  =  42875     .     .     (2) ; 
extracting  the  cube  root  of  both  members  of  (2), 

35 

a:?/  =  35  ;         .  • .      y  =  —  '» 

substituting  and  reducing, 

x^  —  2x  =  35, 


X    =\  ±  -y/35  +  1  =  1  ±  6; 
:  7,     and      —  5,     y  =  5.     and      —  7. 


364  APPKNDIX. 

29.  A  sets  out  from  C  towards  D,  and  travels  8  miles  each  day; 
after  he  had  gone  27  miles,  B  sets  out  from  D  towards  C,  and  goes 
each  day  ^  <jf  the  whole  distance  from  D  to  C ;  after  he  had 
travelled  as  many  days  as  he  goes  miles  in  each  day,  he  met  A  • 
what  is  tht'  distance  from  D  to  C  ? 

Let  X  dt^  lote  the  number  of  miles  from  D  to  C. 

J* 
Then,     —-     will  denote  the  number  of  miles  B  travels  per  day, 
'20  r  Ji 

also  the  number  of  days  that  he  travels  ; 

hence,     — r^r       denotes  the  number  of  miles  travelled  by  B, 

27  +  8a:        "         "  "  «  «  A. 

From  the  conditions  of  the  problem, 

a;2  8a; 


) 


400   '  ■   20 

oleariiiii  of  fractions  and  reducing, 

a;2  —  240x  =  —  10800 

.-.     a;  :^  120  ±  -/-  10800  +  14400  =  120  ±  60  ; 
whence,  x  =  60,         x  =  180. 

30.  There  are  three  numbers  ;  the  difTt-rence  of  the  differences  of 
the  1st  and  2d,  and  2d  and  3d,  is  4  ;  their  sum  is  40,  and  their  con 
tinued  product  is  1764  :  what  are  the  numbers  ? 

Let  .r,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

(a;-2/)-(y-2)=:        4     .     .      (1) 
x-^y  +  z=      40     .      .      (2) 
xyz  =  1764     .     .      (8) ; 


ADDITIONAL    EXAMPLES.  -^65 

combining  (1)  and  (2),  eliminating  x  and  2, 

3y  =  36;         .-.     y  =  12; 
substituting  in  (2)  and  (3), 

X  +  2  =    28     .     .     (4) 
xz-Ul  .      (5); 

•  iombining  (4)  and  (5), 

2—1,     or     21  ,         y  =  21,     or     7. 

31.  There  are  three  numbers  in  arithmetical  progression  :  the 
sum  of  their  squares  is  93,  and  if  the  first  be  multiplied  by  3,  the 
second  by  4,  and  the  third  by  5,  the  sum  of  the  products  will  be  GH : 
what  are  the  numbers  ? 

Let  X  denote  the  first  number,  and  y  their  common  difference. 

From   the  conditions  of  the  problem, 

^2  +  (x  +  yf  +  {x-\-  2yy  =  93     .     .     (1) 
3x-\-4:{x-{-y)  +  5{x  +  2y)  =  QQ     .     .     (2); 

performing  indicated  operations  and  reducing, 

3a;2  ^  5y2  +  6a-y  =  93     .     .     (3) 
12ar  +  14y  =  06,     or     6x  -f  7y  =  33     .     .     (4). 

33  -Qx 


From  (4),  y  = 


7 


1089  —  396x  +  36a;3  ,  3Sx  —  6** 

•  •  •     y'  = ^Q -'         and         xy  = : 

substituting  in  (3)  and  reducing, 

198  290 

25  25  ' 


?i66  Al'PKNUIX. 

whence     x  -  ^  4-  ,  /   '296   9801  _  99  ±  49  ^ 
25   V    25  ^  625      25   ' 

148 

Taking  the  second  Vcalue  of  x,  we  find     y  =  3,     and  the  nuuibery 
are  2,  5  and  8. 

The  problem  supposes  the  numbers  entire,  therefore  the  1st  value 
of  X  is  not  used. 

32.  There  are  three  numbers  in  r.iiihmetical  progression  whose 
sum  is  9,  and  the  sum  of  their  feurth  powers  is  353  ;  what  are  the 
numbers  % 

Let  a;,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

-ly^x^z     .      .      (1) 
a;  +  y  +  2=:9     .     .      (2) 
«*  +  y*  +  2*  =  353     .     .     (3). 
From  (1)  and  (2)  we  find  y  =  3  ; 

substituting  in  (2)  and  (3), 

a;    +  s    =      6     .     .      (4) 
a;*  +  2*  =  272     .     .      (5)  ; 
raising  both  members  of  (4)  to  the  4th  power, 

x"  +  4a-32  +  ^x^^  +  4a:z3  +  2*  =  1296     .     .     (G) ; 
adding  equations  (5)  and  (6),  member  to  member,  and  dividing  by  2 

a;*  +  2x^2   ^  3x'z2  +  ^xz^  +  2*  =  784     .     .     (7); 
extracting  the  square  root  of  both  members, 

x2  +  xz  +  £2  =  28     .     .     (8)  ; 


ADUlTIOXAf,    KXAMPLES.  367 

squaring  both  members  uf  (4), 

x-  +  2x2  +  2^  =  36     .     .     :9) ; 

fr«m  (8)  and  (9)  we  find 

a;2  =  8     .     .     (10); 

from  (4)  and  HO)  we  get 

a;  =  2,     or     4 ;  2  =  4,     or     2  : 

hence,  the  numbers  are  2,  3  and  4. 

33.   How  many  terms  of  the  arithmetical  progression   1,  3,  5,  7. 
&c.,  must  be  added  together  to  produce  the  6th  power  of  12? 
The  Gth  power  of  12  is  2985984. 
From  Art.  175  v/e  have  the  formula, 


<;  -  2a  ±   y/id  -  2a)2  -\-  ^d  S 
2d 

\v  the  present  case,         a  =  1,     cZ  =  2,     and     S  =  2985984  ; 


yU)  X  2985984 

TT 


substituting,         n  =  — :^ =  1728. 


34.  The  sum  of  6  numbers  in  arithmetical  progression  is  48  ;  the 
product  of  the  common  difference  by  the  least  term  is  equal  to  the 
number  of  terms  :  what  are  the  terms  of  the  progression  ? 

Let  X  denote  the  1st  term,  and  y  the  common  difference. 

From  the  conditions  of  the  problem, 

6x  4-  15y  =  48,     xy  =  Q',         .    .     y  =  -; 

substituting  and  reducing, 

x^  —  Sx  =  —  \o  ; 


.-.     X  =  4  ±  -/-  15  -h  16  =  4  ±  1,     or     r  =  5,     ar  =  3 ; 
whence,  y  =  h         y  =  2: 


268  APPENDIX. 

hence,  the  series  is       3.5.7.9.11.13, 

5.6i.7|.8f  .DA.ii. 

35.  What  is  the  sum  of  10  square  numbers  whose  square  roots 
are  in  arithmetical  progression  the  least  term  of  which  is  3,  and  the 
common  difference  2? 

Let  X  denote  the  sum. 

The  progression  of  roots  is 

3.5.7.9.11  .  13.  15.  17.  19.21, 
and  the  series  of  squares, 

9  .  25  .  49  .  81  .  121  .  169  .  225  .  289  .  361  .  441. 
1st  order  of  difis,  16,  24,  32,  40,  &c., 

2d  order  of  diffs,  8,  8,  8,  &c., 

3d  order  of  diffs,  0,  0,  dzc. 

From  Art.  210,  making 

S'  =  x,     a  =  9,     »  =  10,     rfj  =  16,     ^2  =  8.     ^^  =  0»     *2C 
a;  =  90  4-  45  X  16  +  120  x  8  i=  1770. 

36.  Three  numbers  are  in  geometrical  progression  whose  sum  is 
9*1,  and  the  sum  of  their  squares  is  3325 :  what  are  the  numlicrs? 

Let  a;,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y"^  =z  xz     .     ,     ( I ) 
3;2  4- y2  +  22  =  3325     .     .     (2) 

a;    -I-  y    +  2    =       95     .     .     (3) ; 
combining  (1)  and  (2), 

x^  +  2x2   I-  02  _  3325  -f-  a-2     .     .     (4)  ; 


ADDITIONAL    EXAMPLES.  269 

conibining  (1)  and  (3), 


X  +  ^/xz  +  z  —  9b     .     .     (5)  ; 


from  (4)  and  (5), 


X  +  2  =  ^3325  +  X2 
X  -\-  z  —  95  —  -y/xz ; 


hence,  -v/3325  +  xz  r=  95  —  ^/xz\ 

squaring  both  members, 

3325  +  0-0  =  9025  -  190  ^/xz  +  xz ; 

.  • .      v^  =  30,  or     xz  =r  900     .     .     (6)  ; 

substituting  in   (5),  a;  +  2  =    65     .     .     (7)  • 

from  (6)  and  (7),         a-  =  20  and     45, 

y  =  45  and     20. 

37.  Three  numbers  are  in  geometrical  progression  :  the  difference 
of  the  first  and  second  is  6;  that  of  the  second  and  third  is  15 
what  are  the  numbers  \ 

Let  a:,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y"^  =  xz     .     .     (I; 
a;  —  y=—    6;         .'.     a:  =  y—    6 
y  —  2==— 15;         .-.     2  =  y  +  15, 
and  xz  :=  y"^  +  9y  —  90; 

substituting  in  (1)  we  find         y  =  10; 

.  • ,     a;  =  4,     and     z  =  25. 

38.  There  are  three  numbers  in  geometrical  progression  ;  the  sum 


270  APPENDIX. 

of  the  first  and  r>econd  is  14,  and  the  ditlerence  of  the  second  and 
third  is  15  :   what  are  the  numbers? 

Let  X,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 
y2  —  .rs     .     .     (1 ) 
a:  +  y  =  14;  .*.     a;  =  14  —  y 

z  —y=\h\         .-.     2  =  15  +  y, 
and  a;2  —  210  —  y  —  y2  J 

substituting  in  (1 ),  y^  _|_  |  _  105  ^ 


1  /     ,         1         -  1  ±41       ,^  ,  21 

taking  the  1st  value  of  y,  we  find 

x  =  4,         z  =  25. 

39.  A,  B  and  C  purchase  coffee,  sugar  and  tea  at  the  same  prices 
A  i>ays  $11,62^  for  1\  pounds  of  coffee,  3  ponnds  of  sugar,  and  2i 
pounds  of  tea ;  B  pays  $16,25  for  0  pounds  of  coffee,  7  pounds  of 
sugar,  and  3  pounds  of  tea;  C  pays  $12,25  for  2  pounds  of  coffee, 
54  pounds  of  sugar,  and  4  pounds  of  tea :  what  is  the  price  of  a 
pound  of  each  % 

Let  x,  y  and  z  denote  the  number  of  cents  that  the  coffee,  sugar 
and  tea  cost,  respectively. 

From  the  conditions  of  the  problem, 

7^x+    3y  +  2iz  =  1162i  •  •  (1) 

9a-+    7y  +  32=  1625  •  •  (2) 

2  X  +  5ly  +  4  2  =  1?25  .  .  (3) ; 


ADDITIONAL    KXAMPLE8.  271 

clearing  (1)  and  (3)  of  fractions, 

30a;  +  12y  +  90    =  4650       •     •     (4) 
4  X  -\-  \ly  +  Sz    =2450       •     •     (5). 

From   (2)   and    (4), 

3x-9y  =  -  225,     or     x  -Si/=  —7b     •     •     (6) ; 

from   (2)  and   (5), 

GOa;  +  23y  =  5650     •     •     (7)  ; 

from   (6)  and   (7),  y  =  50 ; 

hy  substitution,  x  =  75,  z  =  200. 

40.   Divide  100  into  2  such  parts  that  the  sum  of  their  square 
roots  shall  be  14. 

Let  x  denote  the  first  part. 

From  the  conditions  of  the  problem, 

V^+  -/lOO-a;=:  14; 
squaring  both  members  and  reducing. 


^100,r-a;2  =  48; 

squaring  both  members  and  reducing, 

a;2-  100.r  =  —2304; 


.-.     a;  =  50  ±  -/—  2304  -\-  2500  =  50  ±  14, 
X  z=  64,     and     36. 

41.  In  a  certain  company  there  were  three  times  as  many  gentle- 
men as  ladies  ;  but  afterwards  8  gentlemen  with  their  wives  went 
away,  and  there  then  remained  five  times  as  many  gentlemen  as 
ladies :    how   many  gentlemen,  and   how  many    ladies    xere  there 

originally  ? 

12 


273  APPKNDIX. 

Let  Sx  denote  the  number  of  gentlemen  ;  then  will  x  denote  the 
number  of  ladies. 

From  the  conditions  of  the  problem, 

Sx-S  =  5{x  —  S)', 

.'.     X  =  ]6,     and     3x  =  48. 

42.  Find  two  quantities  such  that  their  sum,  product,  and   the 
difference  of  their  squares,  shall  all  be  equal  to  each  other. 

Let  X  and  y  denote  the  quantities. 
From  the  conditions  of  the  problem. 

X   +  y    =  xy     •     .     (1) 

x^  —  y^  =  xy     ■     '     (2)  ; 

by  division  of  (2)  by  (1),  we  have 

X  —  y  =  I,     or     X  =  y  +  I  ; 
substituting  in  (1), 

2y  +  1  =  y^  +  y,     or     y"^  ~  y  =  I  ; 


1           /,    ,    I                       1  ±  a/5 
whence,  y  =  -±y'l+-,     or     y  = ^ ; 

hence.  .  =  ^^- 

43.  A  bought  120  pounds  of  pepper,  and  as  many  pounds  of 
ginger,  and  had  one  pound  of  ginger  more  for  a  dollar  than  of 
j>epper;  the  whole  price  of  the  pepper  exceeded  that  of  the  ging«') 
by  6  dollars  :  how  many  pounds  of  pepper,  and  how  many  ut 
singer  had  he  for  a  dollar  1 

Let  X  denote  the  number  of  pounds  of  pepper  for  a  dollar. 


ADDITICNAL    EXAMPLES.  273 

From  the  conditions  of  the  problem, 
120         120 


X  X  +  1 


6,     or     a;2  +  z  =  20 ; 


1  Ur.       1       — 1±0       , 

.-.     a:=  -^±Y/20  +  -^-^ — ;     hence,     x  =  4. 

The  negative  value  does  not  conform  to  the  conditions  of  the 
special  problem. 

44.  Divide  the  number  36  into  3  such  parts  that  the  second  shall 
exceed  the  first  by  4,  and  that  the  sum  of  their  squares  shall  b« 
equal  to  464. 

Let  x,  y  and  z  denote  the  parts. 
From  the  conditions  of  the  problem, 

a:  +  y  +  z  =  36     •     •     (1) 
y  —  X  ^\       •     •     (2) 
X' -\- y-^  +  z^  -  464  .     .      (3) ; 
from  (1),           «2  _^  2a:y  +  y2  =  1296  -  722  +  02     .     .     (4). 
from  (2),  a;2  -  2a;y  +  y2  ^  16         (5^. 

adding  (4)  and  (5),  member  to  member, 

2x2  4.  2y2  ^  1310  _  72^  +  22     .     .     (g) . 
from  (3),  2^2  -f-  2y2  =    928  -    2^2  .     .     .     .     (7) . 

equating  the  second  members  and  reducing, 

22  -  242  =  -  128  ; 


.-.     z=  12  ±  -/--128  +  144  =  12±4; 
hence^  2  =  16,         2  =  8; 

substituting  the  first  value  in  (1), 

ar  +  y  =  20     •     •     (8)  ; 


374  APPKNDIX. 

from  (2)  and  (8),  y  -  12     and     ar  =  8. 

45.  A  gentleman  divided  a  sum  of  money  among  4  persons,  so 
that  what  the  first  received  was  ^  that  received  by  the  other  three  ,* 
what  the  second  received  was  \  that  received  by  the  other  three ; 
what  the  third  received  was  \  that  received  by  the  other  three,  and 
it  was  found  that  the  share  of  the  first  exceeded  that  of  the  last  by 
$14 :  what  did  each  receive,  and  what  was  the  whole  sum  divided  ? 

Let  a;,  y,  z  and  tr  denote  the  number  of  dollars  that  each  received. 
From  the  conditions  of  the  problem, 

2a;  =  y  +  2  -f  ^^     •     •     (1) 

3y  =  ar  -f  z  -f-  w     •     •     (2) 

Az  ■=  X  -\-  y  -\-  w     -     •     (3) 

X  —  w  =  14i     •     •     (4)  ; 
from  (2)  and  (3), 

X  -\-  to  =  Sy  —  z 

X  -{•  w  =z  Az  —  y  ;     whence,     Zy  —  z  =.  4«  —  jr, 
oi  Ay  =  5z,         z  =  |y     •     •     (5) ; 

fiom  (4),  w  =  a;  —  14     •     •     (6)  ; 

Rubstituting  the  values  of  w  and  z  in  (1)  and  (2), 
'2z  =z  y  +  ±y  +  X  -  \A 

Zy  —  X  -\-  ^y  ■\-  X  —  \A;     whence,  by  reduction, 
5ar  —    9y  =  —  70 
10a:-lly=:       70; 
.  •.     a?  =  40,     y  =  30 ;     and  by  substitution,    z  =  24,     w  =  26. 

46.  A  woman  bought  a  certain  number  of  eggs  at  2  for  a  penny. 
and  as  many  more  at  3  for  a  penny,  but  on  selling  them  at  the  rate 


ADDITIONAL    EXAMPLES.  275 

of  5  for  2  pence,  she  lost  4  pence  by  the   bargain  ;  how  many  did 
she  buy  1 

X  X 

I  .et  X  denote  the  number  at  each  price.     Then  will     .^  +  « 

denote  the  number  of  pence  paid,  and     -^ — - — -      will   denote   the 
number  of  pence  received. 

From  the  conditions  of  the  problem, 

X       X       2(x  +  x)        ^  -  .^„ 

2  +  ^=        5  +  4 ;         reducing,         x  =  120. 

47.  Two  travellers  set  out  together  and  travel  in  the  same  direc- 
tion ;  the  first  goes  28  miles  the  first  day,  26  the  second  day,  24  the 
third  day,  and  so  on,  travelling  2  miles  less  each  day  ;  the  second 
travels  uniformly  at  the  rate  of  20  miles  a  day  :  in  how  many  days 
will  they  be  together  again  ? 

Let    X    denote    the    required    number  of  days.      The    distance 
travelled  by  the  first  in  x  days  is 

[(Art.  176),  since     a  =  28,     cf  =  —  2,     and     n  =  x],     denoted  by 

^  X  [5G  -  {x  -  If],     or     29a;  — a;2; 

and  the  distance  travelled   by  the  second   is   denoted   by  20a:: 
hence,  we  have 

29a;  —  a;'-^  -  20a;,     or     a:  =  9. 

48.  A  farmer  sold  to  one  man  30  bushels  of  wheat  and  40  of 
barley,  for  which  he  received  270  shillings.  To  a  second  man  he 
sold  50  bushels  of  wheat  and  30  of  barley,  at  the  same  prices,  and 
received  for  them  340  shillings  :  what  was  the  price  of  eath  ? 

Let  ..   denote  the  number  of  shillings  for  1  bushel  of  wheat, 
and       y       "         "         "  "  "         "  barley. 


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